Counting Rocks! An Introduction to Combinatorics

🧭 Overview

🧠 One-sentence thesis

The handshake problem demonstrates that the same counting question can be solved through multiple approaches—arithmetic reasoning, sequential counting, and visual graph modeling—all yielding the same answer and introducing core combinatorial techniques.

📌 Key points (3–5)

• The problem: five students each shake hands with all others; the goal is to count the total number of handshakes.
• Three solution methods: (1) arithmetic with double-counting correction, (2) sequential counting by student, (3) visual graph representation.
• Graph terminology: vertices represent students, edges represent handshakes, and a complete graph connects every pair of vertices.
• Common confusion: intersections of line segments in the graph drawing are not vertices—only the labeled dots representing students are vertices.
• Why it matters: this simple problem introduces counting techniques, mathematical proof approaches, and graph theory that will be extended to larger, more complex scenarios.

🔢 Three ways to count handshakes

🔢 Arithmetic with double-counting

• Each of 5 students shakes 4 hands, giving 5 times 4 equals 20.
• But each handshake involves 2 students, so we counted every handshake twice.
• Correction: divide 20 by 2 to get 10 handshakes.
• Example: when student a shakes hands with student b, that single handshake is counted once for a and once for b.

📝 Sequential counting by student

• Label the students a, b, c, d, e and count handshakes in order:
  • Student a shakes hands with 4 others.
  • Student b shakes hands with 3 others (excluding a, already counted).
  • Student c shakes 2 more hands (excluding a and b).
  • Student d shakes 1 more hand (with e).
  • Student e does not need to shake any more hands (all already counted).
• Total: 1 plus 2 plus 3 plus 4 equals 10.
• This method avoids double-counting by ensuring each handshake is recorded only once.

🎨 Visual graph model

• Draw a dot for each student and a line segment connecting each pair of dots.
• Counting the line segments gives 10 handshakes.
• This visual tool is called a graph.

📊 Graph terminology and structure

📊 Basic definitions

Vertices: the dots representing the students in the graph.

Edges: the line segments between vertices representing the handshakes.

Complete graph: a graph where every two vertices are connected by an edge.

• The graph shown is called K5, the "complete graph on 5 vertices."
• Don't confuse: intersection points of line segments that are not labeled are not vertices—they should be ignored.

🔗 Why "complete"

• The graph is complete because every pair of students is connected.
• In K5, every one of the 5 vertices has an edge to every other vertex.
• This models the handshake scenario where every student shakes hands with all others.

🧩 What this problem introduces

🧩 Counting problems

• The handshake problem is a prototype for combinatorial counting.
• The excerpt states that similar questions will become more complicated when the number of students increases beyond 5.
• Example: for 101 students, the handshake count involves adding numbers from 1 to 100.

🧩 Mathematical proofs and techniques

The excerpt mentions that the class will cover:

• Basic counting, sets, and binomial coefficients.
• Advanced counting techniques including bijections, recurrence relations, and generating functions.
• Graph theory and optimization.
• Combinatorial geometry, including planar graphs and coloring problems.

🧩 Extensions of the handshake problem

The excerpt lists example questions that build on the handshake scenario:

Topic area	Example question
Basic counting	How many ways can 5 students line up? How many ways can 3 students be chosen for a project?
Counting techniques	How many ways can 10 rocks be distributed among students?
Graph theory	How many phone calls are needed to share news? How many cables to connect 40 buildings? Can you trace the edges of K5 without lifting your pencil?
Combinatorial geometry	Can K5 be redrawn so edges do not intersect? How many regions can 4 students' lines divide a paper into?

🎯 Pedagogical approach

🎯 Group work and communication

• The excerpt emphasizes that developing good communication patterns is more important at this stage than answering problems.
• A group works best if everyone speaks for about the same amount of time and questions are valued as much as answers.
• For unsolved problems, groups should decide if the question is clear or ambiguous, if all needed information is provided, and provide ideas or strategies.

🎯 Problem-solving process

The exercises ask students to:

• Introduce themselves and share why they are taking the class.
• Try to solve some problems.
• Develop strategies for approaching other problems.
• Generate ideas about more advanced problems.
• Write down explanations of answers and thought processes for solved problems.

🧭 Overview

🧠 One-sentence thesis

Three fundamental counting techniques—triangular numbers, factorials, and binomial coefficients—provide powerful methods for solving classical problems about handshakes, orderings, and subset selection.

📌 Key points (3–5)

• Triangular numbers count cumulative sums (like handshakes) using the formula n(n+1)/2.
• Factorials count the number of ways to arrange n distinct objects in order.
• Binomial coefficients count the number of ways to choose k elements from n elements when order doesn't matter.
• Common confusion: choosing vs. arranging—binomial coefficients ignore order (choosing), while factorial-based counts care about order (arranging).
• Two proof strategies: rearrangement proofs (algebraic manipulation) and combinatorial proofs (counting the same thing two different ways).

🔢 Triangular numbers and cumulative sums

🔢 What triangular numbers count

Triangular numbers Tₙ = n(n+1)/2 represent the sum of the first n integers: 1 + 2 + ... + n.

• The name comes from arranging objects in a triangular pattern.
• The first several triangular numbers are 1, 3, 6, 10, 15, ...
• Real-world application: The number of handshakes among n+1 people equals Tₙ.
• Example: For 5 students, handshakes = 1+2+3+4 = 10; for 101 students, handshakes = 1+2+...+100 = 5050.

🧮 Proof by rearrangement

• Strategy: Pair numbers from opposite ends of the sequence.
• Pair (1 + n), (2 + (n-1)), (3 + (n-2)), etc.
• Each pair sums to n+1.
• If n is even: there are n/2 pairs, so total = (n+1) × n/2.
• If n is odd: there are (n-1)/2 pairs plus the middle number (n+1)/2, which also gives n(n+1)/2.

🎨 Combinatorial proof using rocks

• Strategy: Count the same collection of objects in two different ways.
• Arrange n(n+1) rocks in a rectangular grid with n rows and n+1 columns.
• Paint the lower-left half blue: 1 rock in row 1, 2 in row 2, ..., n in row n.
• First count: Blue rocks = 1 + 2 + ... + n.
• Second count: Blue rocks = exactly half of n(n+1) total rocks = n(n+1)/2.
• Since both counts describe the same blue rocks, the formula is proven.

🔄 Factorials and ordering problems

🔄 What factorials count

Factorial n! = n · (n-1) · (n-2) · ... · 3 · 2 · 1 counts the number of ways to arrange n distinct objects in order.

• Definition: 1! = 1, and for n > 1, n! = n · (n-1)!
• Examples: 2! = 2, 3! = 6, 4! = 24, 5! = 120.

📋 The reasoning behind factorials

• Why multiplication works: Each position has a shrinking number of choices.
• Example: 5 students lining up at a desk.
  • First position: 5 choices.
  • Second position: 4 remaining choices.
  • Third position: 3 choices.
  • Fourth position: 2 choices.
  • Fifth position: 1 choice.
  • Total: 5 · 4 · 3 · 2 · 1 = 5! = 120.

🎯 Partial arrangements

• When only some objects are arranged, use a truncated product.
• Example: If only 3 of 5 students line up, the count is 5 · 4 · 3 = 5!/2!.
• Example: 47 students seated in 47 labeled chairs = 47! ways.

🎲 Binomial coefficients and subset selection

🎲 What binomial coefficients count

The binomial coefficient (n choose k), written as C(n,k), is the number of ways to choose k elements from a set of n elements when order does not matter.

• Pronounced "n choose k."
• Formula: C(n,k) = n! / [k!(n-k)!]

🔍 Why order doesn't matter

• Key distinction: Choosing vs. arranging.
• When choosing 2 numbers from {1, 2, ..., n}, selecting {2, 5} is the same as selecting {5, 2}.
• Example: C(n,2) = n(n-1)/2 because there are n choices for the first element and n-1 for the second, then divide by 2 since order doesn't matter.

🧩 Two methods to derive the formula

Method 1: Adjust for overcounting

• Start with ordered selections: n · (n-1) · ... · (n-k+1) = n!/(n-k)!
• The same k-element set can be chosen in k! different orders.
• Divide by k! to remove the ordering: n! / [k!(n-k)!]

Method 2: Line up and select

• There are n! ways to line up all n numbers.
• Choose the first k elements in line.
• Divide by k! because order among chosen numbers doesn't matter.
• Divide by (n-k)! because order among unchosen numbers doesn't matter.

📊 Combining binomial coefficients with factorials

Problem	Setup	Answer	Reasoning
47 students, 50 labeled chairs	Choose chairs, then assign students	C(50,47) · 47!	Choose which 47 chairs to use, then arrange students in those chairs
Same problem	Add 3 ghost students	50! / 3!	Arrange all 50 "students" (including ghosts), then divide by arrangements of ghosts

• Don't confuse: Both answers equal the same value because C(50,47) · 47! = [50!/(47!·3!)] · 47! = 50!/3!

🔑 Proof strategies recap

🔑 Rearrangement proofs

• Manipulate the algebraic expression directly.
• Example: Pairing numbers from opposite ends to show 1+2+...+n = n(n+1)/2.

🔑 Combinatorial proofs

• Count the same collection of objects in two different ways.
• If both methods count the same thing, the two expressions must be equal.
• Example: Counting blue rocks in a rectangular grid proves the triangular number formula.

🧭 Overview

🧠 One-sentence thesis

Graphs are visual tools that represent relationships between objects using vertices and edges, and they help solve problems about connections, walks, and equivalence between different-looking structures.

📌 Key points (3–5)

• What a graph represents: vertices stand for objects (people, places) and edges represent connections (handshakes, bridges).
• Bipartite graphs: useful when modeling connections between two different types of things (e.g., students and employers).
• Walks on graphs: some graphs allow you to trace every edge exactly once without lifting your pencil; others (like Königsberg) do not.
• Common confusion: two graphs that look different in a drawing can actually be the same graph—only which pairs of vertices are connected matters, not positions or shapes.
• How to tell graphs apart: compare number of vertices, number of edges, or other structural properties like degree.

🎨 What graphs are and how to use them

🎨 Basic definition

A graph uses vertices to represent objects and edges to represent connections between pairs of objects.

• The excerpt uses handshakes as an example: each person is a vertex, each handshake is an edge.
• The graph K_n has n vertices, and every pair of vertices is connected by an edge; the number of edges in K_n is "n choose 2."
• Example: K_5 represents 5 students shaking hands with each other.

🔗 Bipartite graphs

A bipartite graph models connections between two different types of things.

• The excerpt gives the example of 5 students meeting 3 employers at a job fair: each employer shakes hands with each student.
• Draw one set of vertices for students and another set for employers; edges only connect students to employers, not within the same group.
• Bipartite graphs are useful for studying relationships like students-and-classes or students-and-employers.

🚶 Walks on graphs

🌉 The Königsberg bridge problem

• The problem asks: can you plan a walk that crosses each of the seven bridges exactly once?
• In 1736, Euler proved it is impossible.
• The excerpt presents a related question: can you trace your pencil along the edges of the Königsberg graph so that each edge is drawn exactly once, without lifting your pencil?

🖊️ Tracing edges

• The challenge is to draw every edge exactly once in one continuous path.
• Whether this is possible depends on the structure of the graph, not on how it is drawn.

🔄 When are two graphs the same?

🔄 What "sameness" means

• When drawing a graph, the position of vertices and edges does not matter.
• The only relevant information is which pairs of vertices are connected.
• Two pictures can look completely different but represent the same graph if the connection pattern is identical.

🔍 How to tell if two graphs are different

The excerpt lists methods:

Method	What to check
Number of vertices	If the counts differ, the graphs are different.
Number of edges	If the counts differ, the graphs are different.
Other structural properties	Degree of vertices, planarity, bipartiteness, etc.

• Don't confuse: two graphs with the same number of vertices and edges can still be different if their connection patterns differ.
• Example: the excerpt asks you to find two such graphs that are still different, and to identify other ways of distinguishing graphs.

📐 Graph properties

📐 Degree of a vertex

The degree of a vertex is the number of edges adjacent to that vertex.

• The excerpt uses the cube graph as an example and asks for the degree of each vertex.
• Degree is a structural property that can help distinguish graphs.

🗺️ Planar graphs

A planar graph can be drawn on a piece of paper with no edges crossing.

• The excerpt asks whether the cube graph is planar.
• Planarity is another way to classify and compare graphs.

🎨 Bipartite property revisited

A graph is bipartite if its vertices can be colored with two colors (e.g., green and gold) so that vertices of the same color are not connected by an edge.

• The excerpt asks whether the cube graph is bipartite.
• This property is useful for problems involving two distinct groups.

🧩 Application examples

🧩 Snow plow operator problem

• Vertices represent apartment buildings, edges represent streets.
• The operator needs to plow each street, return home, and minimize travel.
• The excerpt asks:
  • How to remove edges from the Königsberg graph to make the operator happy.
  • How to add edges to make the operator happy.
  • Whether the location of the operator's house matters.
  • What must be true about a graph for the operator to be happy (in mathematical terms).

📬 Mail carrier problem

• Vertices represent apartment buildings, edges represent streets.
• The carrier needs to deliver mail to each building, return home, and minimize road travel.
• The excerpt asks:
  • Draw graphs where the carrier is happy and where the carrier is unhappy.
  • Whether the location of the carrier's house matters.
  • What must be true about a graph for the carrier to be happy (in mathematical terms).

🌉 Relation to Königsberg

• The excerpt asks how the graph of Königsberg relates to the picture of the bridge problem.
• It also asks how the pencil-tracing question relates to the bridge-crossing question.
• Both involve the same underlying graph structure, just presented in different ways.

🧭 Overview

🧠 One-sentence thesis

SAGE is a free, online, open-source computing tool that allows students to compute combinatorial numbers and expressions quickly and accurately without the slowness and errors of manual calculator work.

📌 Key points (3–5)

• Why use SAGE: Exact formulas for triangular numbers, factorials, binomial coefficients, and other famous numbers can be slow and error-prone on calculators; SAGE is faster and more powerful.
• What SAGE is: A free online open-source program based on Python, accessible through web interfaces without installation.
• How to use it: Type simple commands (e.g., arithmetic, factoring) and evaluate; for larger jobs or saving work, create a free CoCalc account.
• Key advantage over manual methods: SAGE can compute many values at once using list comprehensions and conditional filtering, making exploration much faster.

💻 What SAGE is and where to find it

💻 Definition and access

SAGE: a free online open-source program, based on Python.

• Quick access: Go to http://sagecell.sagemath.org/ for immediate use without an account.
• For larger jobs or saving work: Create a free CoCalc account at http://www.sagemath.org/.
• Documentation: The SAGE reference manual is at http://doc.sagemath.org/html/en/reference/.
• Advanced combinatorics: A tutorial for using SAGE in combinatorics is at https://doc.sagemath.org/html/en/reference/combinat/sage/combinat/tutorial.html.

🧮 Why use computing software instead of calculators

• Speed: Computing many values manually or on a calculator is slow.
• Accuracy: Calculators introduce small errors; SAGE provides exact results.
• Power: SAGE can handle complex expressions and large numbers that are impractical for manual computation.
• Example: Computing factorials of large numbers or powers like 4 to the 4 to the 4 is tedious by hand but instant in SAGE.

🔧 Basic SAGE commands

➕ Simple arithmetic and factoring

• Arithmetic: Type expressions like 2+3 and evaluate.
• Factoring: Use factor(2021) to factor integers.
• Powers: Compute exponents with the caret symbol, e.g., 3^6 gives 729.
• Factorials: Use factorial(6) to compute 6 factorial (which equals 720).

🔍 Comparing values

Example from the excerpt: Which is bigger, 3 to the 6th power or 6 factorial?

• Method 1: Compute each separately:
  • 3^6 outputs 729
  • factorial(6) outputs 720
• Method 2: Use a comparison expression:
  • 3^6 - factorial(6) > 0 outputs True, confirming that 3^6 is larger.

This approach avoids manual comparison and gives a direct true/false answer.

📋 Computing multiple values at once

📋 List comprehensions

SAGE can compute many related values in a single command using list comprehensions.

Example from the excerpt: For which values of k is the binomial coefficient (11 choose k) bigger than 100?

• Slow method: Compute (11 choose k) for each k from 0 to 11 individually and check each result.
• Fast method: Compute all at once:
  • [binomial(11, k) for k in range(12)] outputs [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1]
  • From this list, you can see that (11 choose k) > 100 when k is between 3 and 8.

🎯 Filtering with conditions

You can combine computation with filtering to get only the values that meet a condition.

• Command: [k for k in range(12) if binomial(11,k) >100]
• Output: [3, 4, 5, 6, 7, 8]
• This directly gives the values of k that satisfy the condition, without needing to manually inspect the full list.

⚠️ Important note about range(m)

• The command range(m) includes all integers from 0 up to m minus 1.
• Example: range(12) includes 0, 1, 2, ..., 11 (not 12).
• Don't confuse: range(12) does not include 12 itself; it stops at 11.

🧪 Example problems from the excerpt

🧪 Comparing exponentials and factorials

Problem: Which is bigger, 3^6 or 6!?

• Compute 3^6 → 729
• Compute factorial(6) → 720
• Conclusion: 3^6 is bigger by 9.

🧪 Finding binomial coefficients above a threshold

Problem: For which values of k is (11 choose k) bigger than 100?

• Generate all binomial coefficients: [binomial(11, k) for k in range(12)]
• Inspect the list or filter directly: [k for k in range(12) if binomial(11,k) >100]
• Answer: k = 3, 4, 5, 6, 7, 8.

🧪 Very large computations

The exercises mention computing 4^(4^4) and (4!)! (factorial of 4 factorial).

• These numbers are astronomically large and impractical to compute by hand or on most calculators.
• SAGE handles them instantly, demonstrating its power for combinatorial and number-theoretic exploration.

🧭 Overview

🧠 One-sentence thesis

Mathematics uses different labels—theorem, lemma, proposition, corollary, conjecture, definition, example, and remark—to organize facts by their role and proof status, helping readers understand which statements are proven tools, which are unproven beliefs, and which are simply explanations.

📌 Key points (3–5)

• Proven facts have different labels: theorem, lemma, proposition, and corollary are all proven statements, but they serve different purposes in mathematical exposition.
• Lemmas vs propositions: lemmas are proven as tools to help prove bigger theorems (like subroutines in coding), while propositions are interesting on their own but less central than the main theorem.
• Conjectures vs theorems: a conjecture is a statement mathematicians believe to be true but have not yet proven; once proven, it becomes a theorem.
• Common confusion: definitions require no proof—they simply describe what a term means—while theorems, lemmas, propositions, and corollaries all require rigorous proofs.
• Why labels matter: these tags help readers navigate formal mathematics by signaling whether a statement is a major result, a supporting tool, an immediate consequence, or an unproven hypothesis.

🏛️ Proven statements and their roles

🏛️ What is a theorem

A theorem in mathematics is a statement that is known to be true and has a rigorous proof (or several known proofs).

• Lemmas, propositions, and corollaries are all types of theorems—they differ only in how they are used in mathematical exposition.
• The label "theorem" is typically reserved for important, central results.

🔧 What is a lemma

A lemma is a fact that is proven before proving a big important theorem, because it will be used as a tool in the main proof.

• Why lemmas exist: they break down complex proofs into manageable pieces.
• Analogy: lemmas are like subroutines in coding—reusable building blocks.
• When to use: if the base case of an induction is complicated, it can be proven as a lemma first and then referred to in the main proof.
• Sometimes an important mathematical fact is called a lemma because it is used to prove many different things.

📄 What is a proposition

A proposition is a fact that is interesting on its own, is not just a tool to prove other theorems, but is perhaps not quite as big or interesting as the theorem you're talking about in that section or paper.

• Key distinction: propositions stand alone as interesting results, unlike lemmas which are primarily tools.
• Subjectivity: the excerpt notes "it is indeed all rather subjective"—the choice between calling something a theorem or a proposition depends on the author's judgment of its importance.

➡️ What is a corollary

A corollary is a fact that follows immediately from a theorem that was just stated, perhaps with just one or two lines of proof needed to explain why it follows.

• Immediacy: corollaries require minimal additional work beyond the theorem they follow.
• They capture consequences that are obvious once the main theorem is proven.

🤔 Unproven and non-proof statements

🔮 What is a conjecture

A conjecture is a statement that mathematicians believe to be true but do not have a proof.

• Example from the excerpt: if you notice that the sum of the first 5 odd numbers is 5 squared, and the sum of the first 6 odd numbers is 6 squared, you might conjecture that the sum of the first n odd numbers is always n squared—but until you prove it, it remains a conjecture, not a theorem.
• Famous example: Goldbach's conjecture states that every even number larger than four is the sum of two prime numbers; mathematicians have many such conjectures they don't know how to prove.
• Don't confuse: a conjecture becomes a theorem only after someone provides a rigorous proof.

📖 What is a definition

A definition is a statement that describes what a term means in mathematics.

• No proof needed: definitions require no proof—they are just definitions.
• Sometimes definitions are also referred to as notation.
• Key distinction: unlike theorems, lemmas, propositions, and corollaries, definitions do not make claims that need proving; they simply establish meaning.

🧪 What is an example

An example is an instance of a theorem in a special case, usually obtained by plugging in numbers for the variables in the theorem or considering an application.

• Examples illustrate how a theorem works in practice.
• They help readers understand abstract statements by showing concrete cases.

💡 What is a remark

A remark is an observation whose proof is often clear or brief.

• Purpose after a theorem: remarks often add insight that might not be obvious from simply reading a theorem statement.
• Pointing to more: other times, a remark may point to a stronger theorem which is known to be true, but whose proof is not given in the current exposition.

📊 Summary comparison

Label	Proven?	Purpose
Theorem	Yes	Important, central result with rigorous proof
Lemma	Yes	Tool proven to help prove bigger theorems (like a subroutine)
Proposition	Yes	Interesting on its own, but less central than the main theorem
Corollary	Yes	Follows immediately from a theorem with minimal extra proof
Conjecture	No	Believed true but not yet proven
Definition	N/A	Describes what a term means; no proof needed
Example	N/A	Special case illustrating a theorem
Remark	Often yes, briefly	Adds insight or points to related results

🧭 Overview

🧠 One-sentence thesis

This section uses student dialogue to illustrate common counting mistakes—such as overcounting due to order, undercounting by missing cases, and misapplying probability—and motivates the need to learn systematic counting principles.

📌 Key points (3–5)

• Two example problems: choosing 6 numbers from {1, …, 20} with at least four odd, and counting poker "full house" hands.
• Student a's mistake: multiplies choices in sequence but forgets that order doesn't matter, leading to overcounting.
• Student b's insight and error: correctly recognizes order doesn't matter and uses combinations, but still makes mistakes (e.g., in the first problem, doesn't account for "at least four" meaning four, five, or six odd numbers).
• Student c's confusion: tries to use probability but gets a non-integer answer, revealing a conceptual error.
• Common confusion: when order matters vs. when it doesn't; when to multiply vs. when to add; how to handle "at least" conditions (multiple cases).

🎭 The two counting problems

🎲 Problem 2.1.1: Choosing numbers with odd constraints

Question 2.1.1: How many ways are there to choose 6 numbers from the set {1, …, 20} so that at least four of them are odd?

• The set {1, …, 20} contains 10 odd and 10 even numbers.
• "At least four odd" means four, five, or six odd numbers among the six chosen.
• This is a combination problem (order doesn't matter) with a constraint that requires breaking into cases.

🃏 Problem 2.1.2: Poker full house

Question 2.1.2: In a game of poker, you are dealt five cards from a standard 52-card deck. How many ways can you be dealt a full house?

• A full house is defined as a triple of one number (rank) plus a pair of a different number.
• Example: 6♥, 6♣, 6♦ together with 3♥, 3♠.
• The deck has 13 ranks and 4 suits per rank.

🧑‍🎓 Student a: Overcounting by treating order as mattering

🔢 Approach on Problem 2.1.1

• Student a multiplies: 10 · 9 · 8 · 7 · 16 · 15 = 1,209,600.
• Logic: pick the first odd (10 choices), second odd (9 choices), third odd (8), fourth odd (7), then any two from the remaining 16 numbers (16 · 15).
• Error: this counts each set of six numbers multiple times, once for each order in which they could be chosen.
• Example: choosing {1, 3, 5, 7, 2, 4} is counted differently from {3, 1, 5, 7, 2, 4}, but they are the same set.

🃏 Approach on Problem 2.1.2

• Student a multiplies: 52 · 3 · 48 · 3 = 22,464.
• Logic: pick the first card (52 choices), then two more of that rank (3 ways to choose 2 from 3 remaining suits), then pick another card (48 choices), then one more of that rank (3 ways).
• Error: again, order of dealing doesn't matter in a hand, so this overcounts.
• Student b points out: "That's over counting again because it doesn't matter which card is dealt first."

🧑‍🎓 Student b: Recognizes combinations but misses cases

🔢 Approach on Problem 2.1.1

• Student b calculates: (10 choose 4) · (16 choose 2) = 25,200.
• Logic: choose 4 odd numbers from 10, then choose 2 numbers from the remaining 16.
• Correct insight: order doesn't matter, so use combinations.
• Error: "at least four odd" means four or five or six odd numbers; student b only counted the case of exactly four odd and two non-odd.
• Missing cases: exactly five odd (one even) and exactly six odd (zero even).

🃏 Approach on Problem 2.1.2

• Student b calculates: (13 choose 2) · (4 choose 3) · (4 choose 2) = 1,872.
• Logic: choose 2 ranks from 13, then choose 3 suits for the first rank (the triple) and 2 suits for the second rank (the pair).
• Error: (13 choose 2) treats the two ranks symmetrically, but a full house distinguishes which rank is the triple and which is the pair; the correct count should first choose which rank is the triple (13 choices), then which rank is the pair (12 choices), not (13 choose 2).
• The formula overcounts by a factor related to the symmetry.

🧑‍🎓 Student c: Misapplying probability

🔢 Approach on Problem 2.1.1

• Student c calculates: (20 choose 6) · (1/2)^4 = 2422.5.
• Logic: total ways to choose 6 from 20, times the "probability" that four are odd (treating each number as having probability 1/2 of being odd).
• Error: the answer is not an integer, which is impossible for a counting problem.
• The probability approach is incorrect here because the constraint "at least four odd" is not the same as a binomial probability with independent trials; the numbers are chosen without replacement from a finite set.

🃏 Approach on Problem 2.1.2

• Student c doesn't know what a full house is and asks for clarification.
• This highlights the importance of understanding the problem definition.

🧑‍🎓 Student d: Using external data without understanding

🔢 Approach on Problem 2.1.1

• Student d suggests moving to the next question when student c gets a non-integer answer.
• Correct observation: asks "What if there are more than four odd numbers?"—recognizing that "at least four" includes multiple cases.

🃏 Approach on Problem 2.1.2

• Student d looks up the probability of a full house (0.001441) and the total number of 5-card hands ((52 choose 5) = 2,598,960).
• Multiplies them: 0.001441 · 2,598,960 ≈ 3745.10136.
• Correct insight: the number of full houses should equal the probability times the total number of hands.
• Error: the final answer should be an integer; the decimal suggests either the probability or the calculation is imprecise.
• Student d is close but doesn't verify the reasoning or the exact value.

🧩 Common mistakes and lessons

❌ Order matters vs. order doesn't matter

• Don't confuse: when you care about the sequence (permutations) vs. when you only care about the set (combinations).
• Student a's errors stem from treating unordered problems as ordered.
• Example: choosing {1, 3, 5} is the same as choosing {3, 1, 5} if order doesn't matter.

❌ "At least" means multiple cases

• "At least four odd" means four, five, or six odd numbers.
• Student b only counted one case (exactly four odd).
• Lesson: break "at least" constraints into separate cases and add them.

❌ Probability vs. counting

• Probability formulas (like binomial probability) assume independent trials; counting problems with "choose without replacement" require combinatorial methods.
• Student c's non-integer answer signals a conceptual error.

❌ Symmetry and labeling

• In the full house problem, the triple and the pair are not interchangeable.
• Student b's use of (13 choose 2) treats them symmetrically, which is incorrect.
• Lesson: distinguish labeled vs. unlabeled choices.

📝 Exercises and takeaways

📝 What the exercises ask

1. Identify something correct and something wrong in each student's reasoning.
2. Assess how close the students are to solving each problem and what advice to give.

📝 Takeaway

• The section ends with student c saying "Maybe we need to read more of the book."
• This motivates the rest of Chapter 2, which will introduce systematic counting principles (addition, subtraction, multiplication, division, and inclusion-exclusion) to avoid these common mistakes.
• The dialogue format illustrates real pitfalls: overcounting, undercounting, misapplying probability, and not breaking problems into cases.

🧭 Overview

🧠 One-sentence thesis

The addition and subtraction principles speed up counting by splitting collections into shorter lists or by counting a larger set and removing unwanted items.

📌 Key points (3–5)

• Why these principles matter: Counting one-by-one is always possible for finite sets, but it is slow and error-prone; splitting or subtracting speeds things up.
• Addition principle: count separate sub-collections and add the totals together.
• Subtraction principle: count a larger collection and subtract the unwanted items.
• Common confusion: addition requires separate collections (e.g., things ending in 0 vs. things ending in 1), not overlapping groups.
• When to use which: use addition when you can naturally split items into non-overlapping groups; use subtraction when it's easier to count "everything" and remove what you don't want.

🧮 Why we need smarter counting strategies

🐌 The problem with counting one-by-one

• Counting "by 1's" (adding 1 for every item) always works for finite collections.
• But it is very slow and easy to make mistakes:
  • You might forget some items.
  • You might count some items twice.
• Even using a computer to generate the list can be slow if the list is very long.

🧩 The solution: split or subtract

• Split: break the collection into shorter lists, count each, then add.
• Subtract: count a larger set that includes everything, then remove what you don't want.

➕ The addition principle

📐 Definition and core idea

Addition Principle: The sum a + b counts the total number of things in a collection formed by adding a collection of b things to a collection of a things.

• You have two (or more) separate collections.
• Count each collection separately.
• Add the counts together to get the total.

🔢 Example: two-digit integers ending in 0 or 1

• Goal: count two-digit positive integers ending in 0 or 1.
• Step 1: count those ending in 0 → 9 numbers (one for each tens digit: 10, 20, 30, …, 90).
• Step 2: count those ending in 1 → 9 numbers (11, 21, 31, …, 91).
• Step 3: add them → 9 + 9 = 18 total.
• The two groups do not overlap (a number cannot end in both 0 and 1), so addition is safe.

🍎 Example: apples and bananas

• You have 90 pieces of fruit (apples, bananas, and oranges).
• You want to count apples and bananas together.
• Instead of counting each apple and banana one-by-one, you can count apples, count bananas, then add.
• (The excerpt mentions this scenario to motivate the subtraction principle next.)

➖ The subtraction principle

📐 Definition and core idea

Subtraction Principle: If b things are removed from a collection of a things, then there are a − b things left.

• Count a larger, easier-to-count collection.
• Subtract the count of the items you don't want.
• What remains is the count you need.

🍊 Example: apples and bananas (using subtraction)

• You have 90 pieces of fruit total.
• You can easily see there are 3 oranges.
• So apples and bananas together = 90 − 3 = 87.
• This is faster than counting apples and bananas one-by-one.

🔢 Example: two-digit integers not ending in 9

• Goal: count two-digit positive integers ending in 0, 1, 2, 3, 4, 5, 6, 7, or 8.
• Alternative to addition: instead of adding 9 + 9 + 9 + … (nine times), use subtraction.
• Step 1: count all two-digit positive integers → 90 (from 10 to 99).
• Step 2: count those ending in 9 → 9 (19, 29, …, 99).
• Step 3: subtract → 90 − 9 = 81.
• This is simpler than nine separate additions.

🧷 When to use each principle

➕ Use addition when…

• You can naturally split the collection into non-overlapping groups.
• Each group is easy to count separately.
• Example: counting numbers ending in 0 or 1 (two separate groups).

➖ Use subtraction when…

• It is easier to count a larger set that includes everything.
• The unwanted items are easy to count.
• Example: counting all two-digit numbers, then removing those ending in 9.

⚠️ Don't confuse: overlapping vs. separate

• Addition only works if the sub-collections do not overlap.
• If a number could be in both groups, you would count it twice.
• Example: "ends in 0" and "ends in 1" are separate; "ends in 0" and "is even" overlap (10, 20, … are in both), so simple addition would overcount.

🧭 Overview

🧠 One-sentence thesis

The multiplication and division principles provide systematic ways to count outcomes when choices are made in sequence or when identical groupings must be accounted for, and they explain why formulas like combinations divide by factorial terms.

📌 Key points (3–5)

• Multiplication principle: counts the number of ways to make one choice in a ways and then another choice in b ways, giving a · b total outcomes.
• Division principle (two forms): either counts how many collections result when a things are grouped evenly into size b (a/b collections), or counts the size of each collection when a things are divided into c equal groups (a/c per group).
• Why division matters for combinations: when order doesn't matter, we've overcounted by the number of ways to rearrange items, so we divide to correct for this.
• Common confusion: addition vs multiplication—use addition when choosing "one thing or another" (separate cases); use multiplication when choosing "one thing and then another" (sequential choices).
• Combining principles: real problems often require mixing addition (for cases), multiplication (for sequential choices), and division (for overcounting corrections) in the same solution.

🔢 The multiplication principle

🔢 What it counts

Multiplication Principle: The product a · b counts the number of ways to choose one thing in a ways and then another in b ways.

• It applies when you make sequential choices where the second choice has the same number of options regardless of the first choice.
• The key word is "and then"—you perform one action followed by another.

👕 Example: outfits

• Student k owns 15 shirts and 6 pairs of pants. How many outfits (one shirt and one pair of pants)?
• First choose a shirt (15 ways), then choose pants (6 ways).
• Total: 15 · 6 = 90 possible outfits.

🔢 Example: two-digit integers

• Count two-digit positive integers ending in 0, 4, or 8.
• First digit: 9 ways (1 through 9, since it's a two-digit number).
• Second digit: 3 possibilities (0, 4, or 8).
• Total: 9 · 3 = 27 possibilities.

➗ The division principles

➗ Division Principle 1: grouping into fixed-size collections

Division Principle 1: If a things are grouped evenly into collections of size b, there are a/b collections.

• You know the size of each group and want to find how many groups.
• Example: 90 pieces of fruit, 10 pieces per basket → 90 / 10 = 9 baskets needed.
• Example: 36 students, 4 people per van → 36 / 4 = 9 vans needed.

➗ Division Principle 2: dividing into a fixed number of collections

Division Principle 2: If a things are grouped evenly into c collections of equal size, the number of things in each collection is a/c.

• You know the number of groups and want to find the size of each group.
• Example: 90 pieces of fruit divided evenly into 9 baskets → 90 / 9 = 10 pieces per basket.
• Example: 36 students divided evenly into 9 cars → 36 / 9 = 4 people per car.

🎟️ Example: lotto tickets and overcounting

• A lotto ticket is 5 different numbers from {1, ..., 90}, where order does not matter.
• Overcounting approach: There are 90 · 89 · 88 · 87 · 86 ways to pick 5 different numbers one at a time (multiplication principle).
• But this counts each ticket multiple times—once for each ordering of the same 5 numbers.
• Each set of 5 numbers can be arranged in 5! = 120 different orders.
• So we've counted each ticket 5! times.
• Correction: Divide by 5! to get the true count: (90 · 89 · 88 · 87 · 86) / 5! = 43,949,268.
• This explains why the combination formula divides by the factorial.

🤝 Example: pairing people

🤝 Problem setup

• How many different ways can 10 people be grouped into 5 pairs of two?

🤝 Solution 1: multiplication principle

• Order the people by a predetermined criterion (e.g., age).
• The youngest person has 9 possible partners.
• The youngest remaining has 7 possible partners.
• The youngest remaining has 5 possible partners.
• The youngest remaining has 3 possible partners.
• The last two people are paired together.
• Total: 9 · 7 · 5 · 3 · 1 = 945.

🤝 Solution 2: division principle

• There are 10! ways to arrange 10 people in a line.
• Pair the first and second, third and fourth, etc.
• But this overcounts:
  • There are 5! ways to rearrange the ordering of the 5 pairs (doesn't change who is paired with whom).
  • There are 2 ways to switch the order within each of the 5 pairs (2⁵ total), which also doesn't change pairings.
• Total: 10! / (5! · 2⁵) = 945.

🔀 Combining the principles

🔀 When to use each principle

Principle	When to use	Key word
Addition (a + b)	Choosing one thing or another (separate cases)	"or"
Subtraction (a − b)	Removing b things from a things	"not" / "except"
Multiplication (a · b)	Choosing one thing and then another (sequential choices)	"and then"
Division (a/b)	Grouping evenly or correcting for overcounting	"grouped into" / "order doesn't matter"

🔀 Distinguishing addition vs multiplication

• Addition: making one choice or another (mutually exclusive cases).
• Multiplication: making one choice and then another (successive choices).
• Example: 3 car models and 4 bike models.
  • Buy one vehicle (car or bike): 3 + 4 = 7 ways.
  • Buy one car and one bike: 3 · 4 = 12 ways.

👔 Example: matching outfit colors

• Bob has 5 brown shirts and 4 blue shirts, 3 brown pants and 4 blue pants.
• He wants shirt and pants to have the same color. How many ways?
• Split into cases (addition): brown outfit or blue outfit.
  • Brown case: pick one of 5 brown shirts and then one of 3 brown pants → 5 · 3 = 15 (multiplication).
  • Blue case: pick one of 4 blue shirts and then one of 4 blue pants → 4 · 4 = 16 (multiplication).
• Add the cases: 15 + 16 = 31 total outfits.

🔀 Summary table

Expression	Combinatorial meaning
a + b	Total size of a collection formed by adding b things to a things
a − b	Number of elements left after removing b things from a things
a · b	Number of ways to choose one thing from a things and then one from b things
a/b	If a things can be sorted into collections of size b, then a/b is the number of collections

🧭 Overview

🧠 One-sentence thesis

Solving counting problems requires carefully choosing and combining the four basic principles—addition, subtraction, multiplication, and division—by recognizing whether you are making one choice or another (addition) versus one choice then another (multiplication).

📌 Key points (3–5)

• Four basic principles summarized: addition (total size after adding), subtraction (elements left after removing), multiplication (ways to choose one thing then another), and division (number of collections when sorting).
• Common confusion—addition vs multiplication: use addition when making one choice or another; use multiplication when making one choice then another in succession.
• Combining principles: real problems often require splitting into cases (addition) and then applying multiplication within each case.
• Key difficulty: small errors in choosing the wrong principle at any step can make the entire answer wrong.

📋 The four basic principles

➕ Addition principle

a + b: The total size of a collection formed by adding b things to a things.

• Use addition when you are counting "this or that"—mutually exclusive alternatives.
• Example: 3 car models and 4 bike models; buying one vehicle means 3 + 4 = 7 ways (you pick a car or a bike, not both).

➖ Subtraction principle

a − b: The number of elements left after removing b things from a things.

• Straightforward: start with a, take away b, count what remains.

✖️ Multiplication principle

a · b: The number of ways to choose one thing from a things and then one from b things.

• Use multiplication when you are making one choice then another in succession, and the choices do not depend on earlier choices.
• Example: 3 car models and 4 bike models; buying one car and one bike means 3 · 4 = 12 ways (you pick a car then a bike).

➗ Division principle

a/b: If a things can be sorted into collections of size b, then a/b is the number of collections.

• Used when grouping or partitioning objects into equal-sized sets.

🔀 Distinguishing addition and multiplication

🔀 The "or" vs "then" rule

• Addition: one choice or another—alternatives that don't happen together.
• Multiplication: one choice then another—sequential choices made in succession.
• Don't confuse: the same problem can look similar but require different principles depending on whether you are choosing alternatives or making multiple independent choices.

🧪 Example: vehicles

Question	Principle	Calculation	Reason
Buy one vehicle (car or bike)	Addition	3 + 4 = 7	You choose a car or a bike
Buy one car and one bike	Multiplication	3 · 4 = 12	You choose a car then a bike

🧩 Combining principles in practice

🧩 Strategy: split into cases

• Many problems require breaking into mutually exclusive cases (addition) and then applying multiplication within each case.
• Example: Bob has 5 brown shirts, 4 blue shirts, 3 brown pants, and 4 blue pants. He wants an outfit where shirt and pants match in color.
  • Case 1 (brown outfit): pick one of 5 brown shirts then one of 3 brown pants → 5 · 3 = 15 ways.
  • Case 2 (blue outfit): pick one of 4 blue shirts then one of 4 blue pants → 4 · 4 = 16 ways.
  • Total: brown or blue → 15 + 16 = 31 outfits.

⚠️ Watch for errors

• The excerpt emphasizes: "Small errors in the set-up can make an answer completely wrong."
• At each step, ask: am I choosing alternatives (add) or making sequential choices (multiply)?
• Always identify which principle you are using at each step to avoid mixing them up.

🧭 Overview

🧠 One-sentence thesis

A set with n elements has exactly 2 to the power of n subsets, because forming each subset requires n independent yes-or-no decisions about whether to include each element.

📌 Key points (3–5)

• What a set is: an unordered collection of distinct objects; order does not matter, and elements are not repeated.
• What a subset is: a set contained in a bigger set—every element of the subset is also in the larger set.
• The subset-counting formula: a set with n elements has 2 to the power of n subsets (including the empty set and the set itself).
• Why the formula works: for each of the n elements, you make 2 choices (include it or not), so by the multiplication principle the total is 2 × 2 × … × 2 (n times).
• Common confusion: the empty set is a subset of every set, and every set is a subset of itself—these are often overlooked when listing subsets.

📦 What sets are and how to write them

📦 Basic definition of a set

Set: a collection of distinct objects (in no particular order). The objects in a set are called elements.

• Sets are written inside curly brackets: { }.
• Example: {a, b, c} and {1, 2, 3, 4} are both sets.
• Order does not matter: {a, b, c} is the same set as {b, c, a}.
• Don't confuse: a set is not a sequence or list—rearranging elements does not create a different set.

🔢 Size (cardinality) of a set

Size (or cardinality) of a finite set A (written |A|): the number of elements in A.

• Example: |{2, 4, 5}| = 3.
• Counting problems can be phrased as computing the cardinality of a set.
• The excerpt notes that infinite sets also have cardinality, and there are even different sizes of infinity, but this topic is not covered.

✍️ Set builder notation

Set builder notation: {x ∈ A | x has property P} represents the set of all elements x in the set A that satisfy property P.

• Sometimes written {x | x has property P} if the containing set A is clear.
• Example: {x ∈ {2, 4, 5} | x is even} = {2, 4}.
• Example: the set of all integers whose last digit is zero can be written {10 · k | k is an integer}.
• This notation helps describe large or infinite sets without listing every element.

🌐 Famous sets

The excerpt introduces standard notation for important infinite sets:

Symbol	Name	Description
∅	Empty set	The unique set containing no elements: { }
N	Natural numbers	{0, 1, 2, 3, 4, …} (nonnegative integers; note: 0 is included in this book)
Z	Integers	{…, −2, −1, 0, 1, 2, …}
Q	Rational numbers	`{a/b
R	Real numbers	All real numbers

• Example: 1.73 ∈ Q is a good approximation for √3, but √3 itself is not in Q (proven in Section 5.5).
• Example: π is a real number but not in Q (Lambert proved this in the 1760s).
• Example: the set of all even natural numbers can be written {x ∈ N | x is even} = {0, 2, 4, 6, 8, …} = {2 · k | k ∈ N}.

🔣 Membership notation

• b ∈ A means b is an element of the set A.
• b ∉ A means b is not an element of A.
• Example: 4 ∈ {2, 4, 5} but 3 ∉ {2, 4, 5}.

🧩 Subsets and containment

🧩 What a subset is

Subset: a set A is a subset of a set B (written A ⊆ B) if every element of A is also an element of B.

• Example: {2, 4} ⊆ {2, 3, 4}, but {2, 4, 5} is not a subset of {2, 3, 4}.
• Don't confuse: subset vs element—{2} is a subset of {2, 3, 4}, but 2 is an element of {2, 3, 4}.

🔄 Every set is a subset of itself

• For any set A, A ⊆ A because every element of A is an element of A.
• The empty set ∅ is a subset of every set A because there are no elements in ∅ to violate the condition.

🪆 Nested inclusions

• Example: all natural numbers are integers, all integers are rational numbers, and all rational numbers are real numbers.
• This can be written as a nested sequence: ∅ ⊆ N ⊆ Z ⊆ Q ⊆ R.

🔢 Counting the number of subsets

🎯 The subset-counting problem

• Motivating example: four students say they might go to office hours on Monday. How many different groups might show up?
• This is the same as counting the possible subsets of the set {a, b, c, d} of four students.
• The general question: how many subsets does a set of size n have?

🔍 Small cases to build intuition

The excerpt works through small examples:

Set size n	Set example	All subsets	Number of subsets
1	{a}	∅, {a}	2 = 2¹
2	{a, b}	∅, {a}, {b}, {a, b}	4 = 2²
3	{a, b, c}	∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}	8 = 2³

• Pattern: the number of subsets is a power of 2.
• For a set with 4 elements, we might guess 2 to the power of 4 = 16 subsets.

🧮 The theorem and its proof

Theorem 2.5.16: If S is a set with n elements, then S has 2 to the power of n subsets.

Proof idea:

• Label the elements of S as a₁, a₂, a₃, …, aₙ.
• Picking a subset of S is the same as:
  • First, choosing if a₁ is in the subset or not (2 choices),
  • Next, choosing if a₂ is in the subset or not (2 choices),
  • …
  • Finally, choosing if aₙ is in the subset or not (2 choices).
• At each step i, there are 2 choices (include aᵢ or not).
• Since we make a decision n times, the total number of subsets is 2 × 2 × … × 2 (n times) = 2 to the power of n by the multiplication principle.

🌳 Visualizing the proof

The excerpt provides a tree diagram for n = 3 with S = {a₁, a₂, a₃}:

• At each level, you decide "yes" or "no" for one element.
• The tree has 3 levels (one per element).
• The leaves of the tree correspond to the 8 subsets: ∅, {a₃}, {a₂}, {a₂, a₃}, {a₁}, {a₁, a₃}, {a₁, a₂}, {a₁, a₂, a₃}.

🔍 Edge case: the empty set

• The theorem even works when n = 0.
• The empty set ∅ is the unique set with 0 elements.
• There is 1 = 2 to the power of 0 subset of ∅, namely ∅ itself.

🧠 Abstract perspective

• You can think of forming a new set T whose elements are the subsets of S.
• Then 2 to the power of n is the cardinality of T.
• In other words, the "set of all subsets" has size 2 to the power of n.

🎓 Examples and applications

🎓 Office hours example

• Four students {a, b, c, d} say they might go to office hours.
• Number of different groups that might show up: 2 to the power of 4 = 16.
• This includes the empty set (no one shows up) and the full set (everyone shows up).

🎓 Even-sized subsets

• Students a, b, c say they might show up to office hours.
• Among the 8 possible outcomes, how many have an even number of students?
• Answer: 4, namely ∅, {a, b}, {a, c}, {b, c}.
• (Even cardinality means 0, 2, or 4 students; here the maximum is 3, so 0 or 2.)

🎓 Rocks on a cairn (Exercise 4)

• You have 20 different rocks and want to place a subset on a cairn.
• Requirements: use at least 1 rock and do not use all 20.
• Total subsets: 2 to the power of 20.
• Subtract 2 (the empty set and the full set): 2 to the power of 20 − 2.

🎓 Book donation (Exercise 9)

• You have 6 different books and want to donate a subset to the library.
• Requirements: donate at least 1 book and do not donate all 6.
• Total subsets: 2 to the power of 6.
• Subtract 2 (the empty set and the full set): 2 to the power of 6 − 2 = 64 − 2 = 62.

🎓 Subsets containing a specific element (Exercise 10)

• How many subsets of {1, 2, 3, 4, 5} contain the number 1?
• If 1 must be in the subset, you only decide for the other 4 elements.
• Answer: 2 to the power of 4 = 16.

🧭 Overview

🧠 One-sentence thesis

The addition and subtraction principles from counting can be expressed precisely using set operations—union for addition (when sets don't overlap) and difference for subtraction (when one set is contained in another).

📌 Key points (3–5)

• Union and addition: when two finite sets have no elements in common, the size of their union equals the sum of their sizes.
• Intersection: the intersection of two sets contains only the elements that belong to both sets.
• Difference and subtraction: when set B is a subset of set A, the size of A minus B equals the size of A minus the size of B.
• Common confusion: "or" in set union means "or both," not the exclusive "either-or" meaning in everyday phrases like "soup or salad."
• Complement: the complement of B in A is just another name for A minus B when B is a subset of A.

🔗 Union and intersection operations

🔗 Union definition and meaning

Union of sets A and B is A ∪ B = {x : x ∈ A or x ∈ B}.

• The "or" here means x belongs to A, or to B, or to both.
• This is different from the exclusive "or" in everyday language (like "paper or plastic").
• The union collects all elements that appear in at least one of the sets.
• Example: {2, 4, 5} ∪ {2, 5, 7, 9} = {2, 4, 5, 7, 9}—every element from either set appears once in the union.
• Visually, the union covers the entire region of overlapping circles in a diagram.

🔗 Intersection definition and meaning

Intersection of sets A and B is A ∩ B = {x : x ∈ A and x ∈ B}.

• The intersection contains only elements that belong to both sets simultaneously.
• Example: {2, 4, 5} ∩ {2, 5, 7, 9} = {2, 5}—only 2 and 5 appear in both sets.
• Visually, it is the common region where the circles overlap.
• If two sets have no elements in common, their intersection is the empty set ∅.

➕ The addition principle for sets

➕ When addition works

Addition principle for sets: If A and B are finite sets with no elements in common (that is, A ∩ B = ∅), then the union A ∪ B has size |A| + |B|.

• This is the set-theory version of the counting addition principle.
• The key requirement: the sets must have no overlap (their intersection is empty).
• Why it works: if no element is counted twice, you can simply add the sizes.
• Example: A student has climbed 3 big mountains A = {Longs peak, Pikes peak, Mt. Evans} and 2 small mountains B = {Arthur's rock, Horsetooth}. Since there is no overlap between big and small mountains, the total is |A| + |B| = 3 + 2 = 5 mountains.

➕ What happens when sets overlap

• The addition principle does not apply when A ∩ B ≠ ∅.
• If sets share elements, simply adding |A| + |B| would count the shared elements twice.
• Don't confuse: the union still exists when sets overlap, but its size is not |A| + |B|; you need the Principle of Inclusion-Exclusion (covered in the next section).

➖ Difference, complement, and the subtraction principle

➖ Set difference

Difference of sets A and B is A − B = {x ∈ A | x ∉ B}.

• A − B contains elements that are in A but not in B.
• Order matters: A − B and B − A are generally different.
• Example: {2, 4, 5} − {2, 5, 7, 9} = {4}, but {2, 5, 7, 9} − {2, 4, 5} = {7, 9}.
• If A and B have no overlap (A ∩ B = ∅), then A − B = A.

➖ Complement

Complement of B in A: when B ⊆ A, the complement of B in A is the difference A − B, written as B^c when A is clear from context.

• The complement is just a special name for the difference when one set is contained in the other.
• The "c" stands for complement.
• Context matters: the complement depends on which larger set A you are working within.

➖ The subtraction principle for sets

Subtraction principle for sets: If A and B are finite sets and B ⊆ A, then |A − B| = |A| − |B|.

• This is the set-theory version of the counting subtraction principle.
• The key requirement: B must be a subset of A (every element of B is also in A).
• Why it works: removing |B| elements from A leaves |A| − |B| elements.
• Example: How many numbers in A = {1, …, 25} are not a multiple of 5? Let B = {5, 10, 15, 20, 25} be the multiples of 5. Since B ⊆ A, the answer is |A − B| = 25 − 5 = 20.

➖ When subtraction applies

• The subtraction principle requires B ⊆ A.
• If B is not a subset of A, you cannot use |A| − |B| to find |A − B|.
• Don't confuse: A − B always exists as a set, but the size formula |A − B| = |A| − |B| only holds when B ⊆ A.

🧮 Comparing the principles

Principle	Set operation	Key requirement	Size formula
Addition	Union A ∪ B	A ∩ B = ∅ (no overlap)	|A ∪ B| = |A| + |B|
Subtraction	Difference A − B	B ⊆ A (B is a subset of A)	|A − B| = |A| − |B|

• Both principles translate familiar arithmetic operations into set language.
• Both require specific conditions: no overlap for addition, subset for subtraction.
• When the conditions are not met, the formulas do not hold and you need more advanced techniques (like Inclusion-Exclusion for overlapping unions).

🧭 Overview

🧠 One-sentence thesis

The Principle of Inclusion-Exclusion provides a systematic way to count elements in unions of overlapping sets by adding individual set sizes, subtracting intersections, and (for three or more sets) adding back higher-order intersections to avoid double-counting.

📌 Key points (3–5)

• What the principle solves: counting elements in a union when sets overlap non-trivially (share common elements).
• Core mechanism for two sets: add the sizes of both sets, then subtract the intersection to correct for double-counting.
• Extension to three sets: add all individual sizes, subtract all pairwise intersections, then add back the triple intersection.
• Common confusion: forgetting that elements in intersections get counted multiple times when you simply add set sizes—the principle systematically corrects this.
• Visualization tool: Venn diagrams (overlapping circles) help see which regions are counted how many times.

🎯 The two-set principle

🎯 Core formula and meaning

Principle of Inclusion-Exclusion for two sets: If A and B are finite sets, then |A ∪ B| = |A| + |B| − |A ∩ B|.

• Why subtract the intersection: when you add |A| + |B|, every element in A ∩ B is counted twice (once in A, once in B).
• Subtracting |A ∩ B| once removes the duplicate count.
• Example: 5 votes for mint, 4 for caramel, total 9 votes—but two students voted for both, so the actual number of students is 5 + 4 − 2 = 7.

📐 Venn diagram visualization

• Draw two overlapping circles, one for A and one for B.
• The overlap region represents A ∩ B.
• The union A ∪ B covers all parts of both circles.
• Visually, adding |A| and |B| counts the overlap twice; the formula corrects this.

🔢 Worked example: multiples of 3 or 7

• Setup: S = {1, …, 210}; A = multiples of 3 in S; B = multiples of 7 in S.
• |A| = 70, |B| = 30, |A ∩ B| = 10 (multiples of 21).
• |A ∪ B| = 70 + 30 − 10 = 90.
• Finding the complement: numbers with no factor in common with 21 = |S − (A ∪ B)| = 210 − 90 = 120.
• Don't confuse: the complement counts what is not in the union; use the subtraction principle after finding the union size.

🔺 The three-set principle

🔺 Core formula and structure

Principle of Inclusion-Exclusion for three sets: If A, B, C are finite sets, then |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |A ∩ C| + |A ∩ B ∩ C|.

• Pattern: add all individual sizes, subtract all pairwise intersections, add back the triple intersection.
• Why add back the triple intersection: elements in A ∩ B ∩ C are counted three times in the first sum, subtracted three times in the pairwise subtractions (net zero), so you must add them back once.

🧩 Step-by-step logic

1. Start by adding |A| + |B| + |C|: every element in exactly one set is counted once; elements in two sets are counted twice; elements in all three are counted three times.
2. Subtract |A ∩ B|, |B ∩ C|, |A ∩ C|: this removes one count from each pairwise overlap, but elements in A ∩ B ∩ C are subtracted three times (now net count is zero for them).
3. Add |A ∩ B ∩ C| once: restores the count for elements in all three sets to exactly one.

🍓 Worked example: ice cream flavors

• Setup: 7 students; 5 like mint, 4 like caramel, 2 like strawberry; 2 students like both mint and caramel; 1 likes mint and strawberry but not caramel; 0 like strawberry and caramel but not mint.
• Fill in Venn diagram regions step by step.
• Conclusion: exactly 1 student likes all three flavors (the only way to account for the 2 strawberry votes given the constraints).

🔢 Worked example: multiples of 2, 3, or 5

• Setup: S = {1, …, 210}; A = multiples of 2, B = multiples of 3, C = multiples of 5.
• |A| = 105, |B| = 70, |C| = 42.
• Pairwise intersections (using relative primality): |A ∩ B| = 35 (multiples of 6), |A ∩ C| = 21 (multiples of 10), |B ∩ C| = 14 (multiples of 15).
• Triple intersection: |A ∩ B ∩ C| = 7 (multiples of 30).
• |A ∪ B ∪ C| = 105 + 70 + 42 − 35 − 14 − 21 + 7 = 154.
• Finding the complement: numbers with no factor in common with 30 = |S − (A ∪ B ∪ C)| = 210 − 154 = 56.

🖼️ Using Venn diagrams

🖼️ What a Venn diagram shows

• Each set is drawn as a circle (or other closed shape).
• Overlapping regions represent intersections.
• The entire area covered by all circles is the union.
• Helpful for visualizing which elements are counted in which parts of the formula.

🧮 Filling in a Venn diagram

• Start with known totals and constraints.
• Work from the most specific regions (e.g., the triple intersection) outward.
• Use subtraction to find regions: e.g., "likes mint and strawberry but not caramel" = (mint ∩ strawberry) − (mint ∩ strawberry ∩ caramel).
• Example: in the ice cream problem, knowing 1 student likes mint and strawberry but not caramel, and 2 total like strawberry, you can deduce the remaining strawberry region.

⚠️ Common pitfall

• Don't confuse "likes A or B" (union) with "likes A and B" (intersection).
• The principle counts the union; intersections are subtracted to avoid double-counting.

🔗 Relationship to complements

🔗 Counting what is not in the union

• Often the goal is to count elements that satisfy none of several conditions.
• This is the complement of the union: |S − (A ∪ B ∪ C)|.
• Use the subtraction principle: |S − (A ∪ B ∪ C)| = |S| − |A ∪ B ∪ C|.
• First apply inclusion-exclusion to find |A ∪ B ∪ C|, then subtract from the total.

🔢 Example pattern

• "Numbers with no factor in common with 30" = numbers not divisible by 2, 3, or 5.
• Let A, B, C be multiples of 2, 3, 5 respectively.
• Compute |A ∪ B ∪ C| using inclusion-exclusion.
• Answer = |S| − |A ∪ B ∪ C|.

📊 Summary table

Number of sets	Formula structure	Key idea
Two (A, B)	|A| + |B| − |A ∩ B|	Add sizes, subtract overlap once
Three (A, B, C)	|A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |A ∩ C| + |A ∩ B ∩ C|	Add sizes, subtract pairwise overlaps, add back triple overlap
Complement	|S| − |A ∪ B ∪ …|	Use inclusion-exclusion first, then subtract from total

🧭 Overview

🧠 One-sentence thesis

The Cartesian product of two sets formalizes the multiplication principle by showing that the number of ordered pairs from sets A and B equals the product of their sizes.

📌 Key points

• Cartesian product definition: the set of all ordered pairs (a, b) where a comes from set A and b comes from set B.
• Multiplication principle for sets: if A has |A| elements and B has |B| elements, then A × B has |A| · |B| elements.
• Ordered pairs matter: the Cartesian product produces ordered pairs, not just combinations—order is significant.
• Common confusion: don't confuse the Cartesian product (all ordered pairs) with simply listing elements from both sets; the product creates pairs of elements.
• Real-world application: counting problems can be reframed as finding the size of a Cartesian product of smaller, easier-to-count sets.

🔢 Cartesian product fundamentals

🔢 Definition and notation

Cartesian product: If A and B are sets, then their Cartesian product, denoted A × B, is the set consisting of all ordered pairs of elements from A and B: A × B = {(a, b) | a ∈ A and b ∈ B}.

• The product is a new set whose elements are ordered pairs.
• Each pair has a first component from A and a second component from B.
• The notation (a, b) emphasizes that order matters: (a, b) is different from (b, a) unless a = b.

📐 Simple numeric example

The excerpt gives {2, 3, 4} × {x, y}:

• Result: {(2, x), (2, y), (3, x), (3, y), (4, x), (4, y)}
• Each element from the first set pairs with each element from the second set.
• Total pairs: 3 elements × 2 elements = 6 ordered pairs.

Don't confuse: the Cartesian product is not {2, 3, 4, x, y}; it is a set of pairs, not a union of elements.

🗺️ Geometric interpretation

The excerpt mentions R × R (the product of the real numbers with themselves):

• This represents all points (x, y) in the usual coordinate plane R².
• Called the Cartesian plane, which is where the term "Cartesian product" originates.
• Example: [0, 1] × [0, 1] represents all points in the unit square (x, y) where both coordinates are between 0 and 1.

🧮 The multiplication principle for sets

🧮 The formal statement

Lemma (Multiplication principle for sets): If set A has |A| elements and set B has |B| elements, then set A × B has |A| · |B| elements.

• |A| denotes the size (number of elements) of set A.
• The principle connects set size to multiplication: counting ordered pairs reduces to multiplying the sizes of the component sets.
• This is the set-theoretic foundation for the multiplication principle in combinatorics.

🎯 Why it works

• For each of the |A| choices from A, there are |B| choices from B.
• Total number of distinct ordered pairs: |A| choices × |B| choices = |A| · |B| pairs.
• Example from the excerpt: {2, 3, 4} × {2, 5} has 3 · 2 = 6 elements: {(2, 2), (2, 5), (3, 2), (3, 5), (4, 2), (4, 5)}.

🧩 Applied counting examples

🧩 Granola bar indexing

The excerpt describes three boxes of granola bars, each with five bars stacked:

• Index each bar by (i, j) where i ∈ A = {1, 2, 3} (box number) and j ∈ B = {1, ..., 5} (position in box).
• The set of all bars is A × B.
• Total bars: |A × B| = 3 · 5 = 15.

Key insight: real-world objects can be indexed by Cartesian products when they have multiple independent attributes.

🔢 Odd numbers not divisible by 5

The excerpt finds how many numbers in {1, ..., 40} are odd and not multiples of 5:

• These numbers have last digit in C = {1, 3, 7, 9} (odd, not 5).
• Their first digit (tens place) is in B = {0, 1, 2, 3} (to stay ≤ 40).
• The set A of such numbers can be identified with B × C (via the pair (tens digit, ones digit)).
• Size: |A| = |B| · |C| = 4 · 4 = 16.

Strategy: break a complex counting problem into independent choices (digits, positions, attributes), then use the Cartesian product to count.

🔄 Comparison: products vs. other set operations

Concept	What it produces	Size formula	Example
Cartesian product A × B	Set of ordered pairs (a, b)	|A| · |B|	{2, 3} × {x, y} = {(2,x), (2,y), (3,x), (3,y)}
Union A ∪ B (not in excerpt)	Elements in A or B	(not directly multiplicative)	Not covered here
Intersection A ∩ B (not in excerpt)	Elements in both A and B	(not directly multiplicative)	Not covered here

Don't confuse: the Cartesian product creates new elements (pairs) from the originals; it does not simply combine or filter existing elements.

🧭 Overview

🧠 One-sentence thesis

Set partitions organize a set into disjoint blocks, enabling the division principle to count arrangements and equivalence relations to group elements systematically.

📌 Key points (3–5)

• Set partition structure: a collection of disjoint blocks whose union is the entire set.
• Division principle: when a set is partitioned into equal-sized blocks, the number of blocks equals the set size divided by block size.
• Equivalence relations: three properties (reflexive, symmetric, transitive) that naturally create set partitions.
• Common confusion: distinguishing when elements are "the same" (equivalent) vs. "different"—the COOL anagram example shows how treating identical letters as distinct leads to overcounting.
• Modular arithmetic application: congruence modulo m partitions integers into m blocks based on remainders.

🧱 Set partitions and blocks

🧱 What is a set partition

Set partition: a collection of blocks {B₁, B₂, ..., Bₖ} of a finite set A such that (1) the blocks are disjoint (Bᵢ ∩ Bⱼ = ∅ for all i ≠ j), and (2) the union of the blocks is A (B₁ ∪ B₂ ∪ ... ∪ Bₖ = A).

• A block is a non-empty subset of A.
• "Disjoint" means no element appears in more than one block.
• "Union is A" means every element of A appears in exactly one block.
• Example: {{1, 3, 4}, {2, 6}, {5, 7}} is a set partition of {1, 2, 3, 4, 5, 6, 7}.

🔢 The division principle for sets

Division principle: If {B₁, ..., Bₖ} is a set partition of A where each block has the same cardinality m, then |A|/m = k and |A|/k = m.

• This applies only when all blocks have equal size.
• It connects three quantities: total set size, block size, and number of blocks.
• Knowing any two lets you compute the third.

📝 Counting with equal blocks: the COOL example

Problem: How many ways can the letters in "COOL" be rearranged?

• If we treat the two O's as different (O₁ and O₂), there are 4! = 24 arrangements.
• But the two O's are actually identical, so arrangements differing only by swapping O₁ and O₂ look the same.
• We partition the 24 arrangements into blocks of size 2, where each block contains pairs like {CLO₁O₂, CLO₂O₁}.
• By the division principle: 24/2 = 12 distinct rearrangements.
• Don't confuse: the 24 counts distinguishable arrangements when O's are labeled; the 12 counts actual distinct arrangements when O's are indistinguishable.

🔗 Equivalence relations

🔗 Definition and three properties

Equivalence relation: a way of identifying elements x ∼ y with three rules: (1) reflexive: x ∼ x; (2) symmetric: if x ∼ y then y ∼ x; (3) transitive: if x ∼ y and y ∼ z then x ∼ z.

• Reflexive: every element is equivalent to itself.
• Symmetric: equivalence works both ways.
• Transitive: equivalence chains together.
• These three properties ensure that equivalence relations naturally create set partitions.

🎡 Round table seating example

Problem: How many ways can students A, B, C, D be seated at a rotating round table?

• There are 4! = 24 ways to seat them if the table were fixed.
• Two arrangements are equivalent if one is a rotation of the other.
• Example: ABCD ∼ DABC ∼ CDAB ∼ BCDA (all look the same after rotation).
• Each arrangement is equivalent to 3 others, so blocks have size 4.
• By the division principle: 24/4 = 6 distinct seating arrangements.

🔢 Modular arithmetic and congruence

🔢 Congruence modulo m

Congruence modulo m: integers x and y are congruent modulo m (written x ≡ y mod m) when m divides y − x.

• This means x and y have the same remainder when divided by m.
• Key lemma: If r is the remainder when x is divided by m, then x ≡ r mod m.
• Special case: if m divides x, then x ≡ 0 mod m.

🕐 Everyday examples of congruence

Context	Modulus	Example
Clock arithmetic	12	11 + 2 = 13 ≡ 1 mod 12 (1 o'clock after waiting 2 hours from 11 o'clock)
Even/odd	2	Every even integer ≡ 0 mod 2; every odd integer ≡ 1 mod 2
Last digit	10	The last digit of a number indicates its congruence modulo 10

📊 Counting with congruence blocks

Problem: How many numbers in S = {1, ..., 40} are congruent to 1 modulo 5?

• List them: {1, 6, 11, 16, 21, 26, 31, 36}.
• Answer: 8 numbers.
• General pattern: S can be partitioned into 5 blocks (one for each remainder 0, 1, 2, 3, 4), each of size 8.
• Two numbers are in the same block exactly when they are congruent modulo 5.

🧮 General lemma for modular partitions

Lemma 2.9.13: If N is a multiple of m, then S = {1, ..., N} has a set partition into m blocks, each of size N/m, where two numbers are in the same block exactly when they are congruent modulo m.

• This works because every integer has exactly one remainder when divided by m (from 0 to m−1).
• The m blocks correspond to the m possible remainders.
• Each block has equal size N/m because N is a multiple of m.
• Example: For N = 35 and m = 5, there are 5 blocks of size 7 each.

🧭 Overview

🧠 One-sentence thesis

This problem set applies counting principles, set operations, modular arithmetic, and binary string analysis to reinforce the combinatorial and number-theoretic techniques introduced in Chapter 2.

📌 Key points (3–5)

• Counting techniques: multiplication and division principles are used to count handshakes, clothing combinations, and sequences.
• Set operations: problems use cardinality formulas (inclusion-exclusion), Venn diagrams, and complement operations to verify identities and count elements.
• Number theory applications: Euler's totient function counts numbers relatively prime to a given integer; modular arithmetic classifies integers.
• Binary strings and symmetry: binary strings of length m correspond to corners of geometric shapes; complement operations reveal structural symmetries.
• Common confusion: distinguishing between "with repetition" vs "without repetition" in counting problems, and understanding when complement sets equal the original set (depends on parity).

🤝 Counting with multiplication and division principles

🤝 Handshakes between people

• Problem 1: Count handshakes among 7 people if every pair shakes hands exactly once.
• Problem 2: Generalize to n people; prove the count is n times (n minus 1) divided by 2.
• Why division: Each handshake involves two people, so counting "person by person" double-counts every handshake; divide by 2 to correct.
• Example: With 3 people (A, B, C), pairs are AB, AC, BC → 3 times 2 divided by 2 equals 3 handshakes.

👕 Clothing combinations

• Problem 3: Bob has 4 pairs of socks, 2 pairs of shoes, 5 pairs of pants, 10 shirts, and 1 hat. Count ways to get dressed (one of each type).
• Multiplication principle: Multiply the number of choices for each independent item: 4 × 2 × 5 × 10 × 1.
• Don't confuse: This assumes order doesn't matter within each category; we're choosing one item per category, not arranging them.

🔢 Even-digit integers

• Problem 4: How many three-digit positive integers have only even digits?
• Even digits: 0, 2, 4, 6, 8.
• First digit cannot be 0 (must be positive three-digit), so 4 choices; second and third digits each have 5 choices.
• Total: 4 × 5 × 5.

📝 Letter sequences

• Problem 12: Count sequences of 5 letters from A, E, I, O, U, M, S, allowing repeats (e.g., MMSSU).
• 7 letters available, 5 positions, repetition allowed → 7 to the power 5.
• Problem 13: Count sequences with exactly 2 vowels (A, E, I, O, U are vowels; M, S are consonants).
• Choose 2 positions out of 5 for vowels; fill those with 5 choices each; fill remaining 3 with 2 choices each.
• Calculation involves binomial coefficient for positions and multiplication for letter choices.

🧑‍💼 Job assignments

• Problem 14: Assign 5 jobs to 4 people so each person gets at least one job.
• This is a surjective function problem: every person must receive at least one job.
• Use inclusion-exclusion or Stirling numbers of the second kind multiplied by permutations.

🔤 Rearrangements with repetition

• Problem 15: Rearrange the letters of PUPPY (including the original).
• PUPPY has 5 letters: P appears 3 times, U once, Y once.
• Formula: 5 factorial divided by (3 factorial times 1 factorial times 1 factorial).
• Don't confuse: This is a permutation with repetition, not a simple permutation.

🧮 Set operations and cardinality

🧮 Inclusion-exclusion principle

• Problem 5: Given sets A and B with |A| = 21, |B| = 15, and |A ∪ B| = 30, find |A ∩ B|.
• Formula: |A ∪ B| = |A| + |B| − |A ∩ B|.
• Rearrange: |A ∩ B| = |A| + |B| − |A ∪ B| = 21 + 15 − 30 = 6.

🖼️ Venn diagram identities

• Problem 6: Prove A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) using a Venn diagram.
• This is the distributive law for intersection over union.
• Shade both sides on separate diagrams; they match.
• Problem 7: Prove A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) using a Venn diagram.
• This is the distributive law for union over intersection.
• Again, shade both sides; regions coincide.

🔢 Number theory and modular arithmetic

🔢 Euler's totient function

• Problem 8: Count numbers in {1, ..., 105} that are relatively prime to 105.
• A number is relatively prime to 105 if it shares no prime factors with 105.
• Use Euler's totient function φ(n); for n = product of distinct primes, φ(n) = n times (1 − 1/p₁) times (1 − 1/p₂) ...
• Problem 9: For n = pq (p, q distinct primes), show there are pq − p − q + 1 numbers in {1, ..., n} relatively prime to n.
• Numbers not relatively prime: multiples of p or q.
• Use inclusion-exclusion: total n, subtract multiples of p (q many), subtract multiples of q (p many), add back multiples of pq (1 many).
• Result: pq − q − p + 1.

🔢 Modular arithmetic classification

• Problem 16: In {1, 2, ..., 24}, find integers that are 2 mod 8 and compute their sum.
• Numbers: 2, 10, 18 (each is 2 more than a multiple of 8).
• Sum: 2 + 10 + 18 = 30.

🧊 Binary strings and geometric symmetry

🧊 Binary strings of length m

• Problem 10(a): Count binary strings of length m.
• Each position has 2 choices (0 or 1) → 2 to the power m.

🔄 Complement operation

• Problem 10(b): For a string s, the complement sᶜ swaps 0 ↔ 1.
• Example for m = 3: strings are 000, 001, 010, 011, 100, 101, 110, 111.
• Pairs: 000 ↔ 111, 001 ↔ 110, 010 ↔ 101, 011 ↔ 100.

🧊 Geometric interpretation

• Problem 10(c): Plot each binary string s = s₁s₂s₃ as the point (s₁, s₂, s₃) in 3D space.
• The 8 points are the corners of a unit cube.
• Complement pairs are opposite corners (diagonal through the cube's center).

🧊 Subsets and their complements

• Problem 11(a): For m = 2, list all subsets of S (the set of binary strings of length 2) of size k = 3.
• S = {00, 01, 10, 11}; subsets of size 3: {00, 01, 10}, {00, 01, 11}, {00, 10, 11}, {01, 10, 11}.
• Problem 11(b): For a subset T of S containing k strings, Tᶜ is the subset of their k complements.
• Match each subset with its complement subset.

🔄 Parity and self-complementarity

• Problem 11(c): If k is odd, prove T ≠ Tᶜ for any subset T of size k.
• If T = Tᶜ, then every string in T has its complement also in T → strings come in pairs → |T| must be even.
• Contradiction when k is odd.
• Problem 11(d): If k is even, construct T such that T = Tᶜ.
• Take a subset R of size k/2; let T = R ∪ Rᶜ.
• Then T has size k and T = Tᶜ (T is closed under complement).
• Don't confuse: Self-complementarity is possible only when k is even.

🔍 Lemma verification

🔍 Problem 9 (Chapter 2.9)

• Task: Explain why Lemma 2.9.13 is true using Lemma 2.9.3.
• The excerpt does not provide the statements of these lemmas, so the problem requires referring back to earlier sections of Chapter 2.
• This is a proof-writing exercise linking two results within the chapter.

🧭 Overview

🧠 One-sentence thesis

Divisor functions—including the count of divisors, their sum, and the Möbius function—can be computed through formulas based on the prime factorization of a positive integer, and these functions share the multiplicative property when applied to relatively prime integers.

📌 Key points (3–5)

• Divisor functions depend on prime factorization: the value at N depends on how N breaks down into prime factors.
• Three main divisor functions: sigma-zero (count of divisors), sigma-one (sum of divisors), and the Möbius function (alternating sign based on distinct prime factors).
• Multiplicative property: all three functions satisfy f(N·M) = f(N)·f(M) when N and M share no prime factors (relatively prime).
• Common confusion: the multiplicative property only holds when the integers are relatively prime; it fails when they share common prime factors.
• Building formulas progressively: start with prime powers, then products of distinct primes, then general prime factorizations.

🔢 The number of divisors function

🔢 Definition and basic examples

sigma-zero(N): the number of positive divisors of N, including 1 and N itself.

• For N = 22, the divisors are 1, 2, 11, 22, so sigma-zero(22) = 4.
• For N = 23 (a prime), the divisors are only 1 and 23, so sigma-zero(23) = 2.
• For N = 24, the divisors are 1, 2, 3, 4, 6, 8, 12, 24, so sigma-zero(24) = 8.

📐 Formula for prime powers

When N = p raised to power e (where p is prime):

• The divisors are 1, p, p-squared, up to p to the e-th power.
• Formula: sigma-zero(N) = e + 1.
• Example: for N = 81 = 3 to the 4th power, sigma-zero(81) = 5 (the divisors are 1, 3, 9, 27, 81).
• Don't confuse: this formula requires p to be prime; it does not work for composite bases.

🧮 Formula for products of distinct primes

When N = p·q where p and q are distinct primes:

• Formula: sigma-zero(N) = 4 (specifically, 2 times 2).
• The divisors are: 1, p, q, and p·q.
• Example: N = 21 = 3·7 has divisors 1, 3, 7, 21, so sigma-zero(21) = 4.

When N = p·q·r where p, q, r are distinct primes:

• Formula: sigma-zero(N) = 2 times 2 times 2 = 8.
• Each prime can either appear or not appear in a divisor.

🎯 General formula for any factorization

When N = p-one to the e-one power · p-two to the e-two power · ... · p-n to the e-n power (distinct primes):

• Formula: sigma-zero(N) = (e-one + 1) · (e-two + 1) · ... · (e-n + 1).
• Why it works: each divisor is formed by choosing an exponent from 0 to e-i for each prime p-i; the choices are independent and multiply by the counting principle.

➕ The sum of divisors function

➕ Definition and basic examples

sigma-one(N): the sum of the positive divisors of N, including 1 and N itself.

• For N = 22: 1 + 2 + 11 + 22 = 36, so sigma-one(22) = 36.
• For N = 23: 1 + 23 = 24, so sigma-one(23) = 24.
• For N = 24: 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60, so sigma-one(24) = 60.

📊 Formula for prime powers

When N = p to the e-th power (p is prime):

• The divisors are 1, p, p-squared, ..., p to the e-th power.
• Formula: sigma-one(N) = 1 + p + p-squared + ... + p to the e-th power (a geometric series).
• This is the sum of a geometric progression.

🔗 Formula for products of primes

When N = p-one · p-two (distinct primes):

• Formula: sigma-one(N) = sigma-one(p-one) · sigma-one(p-two).
• The sum factors because every divisor of N is uniquely a product of a divisor of p-one and a divisor of p-two.

When N = p-one to the e-one · p-two to the e-two · ... · p-n to the e-n (distinct primes):

• Formula: sigma-one(N) = sigma-one(p-one to the e-one) · sigma-one(p-two to the e-two) · ... · sigma-one(p-n to the e-n).
• Each factor is the geometric sum for that prime power.

🔄 The Möbius function

🔄 Definition and rules

The Möbius function mu(N) is defined by three cases:

• If any prime p-squared divides N, then mu(N) = 0.
• mu(1) = 1.
• If N = p-one · p-two · ... · p-n (distinct primes), then mu(N) = negative-one to the n-th power.

• The function alternates sign based on the number of distinct prime factors.
• It is zero whenever N has a repeated prime factor (not square-free).

🧪 Example and sum property

For N = 22 = 2 · 11:

• Divisors: 1, 2, 11, 22.
• mu(1) = 1, mu(2) = -1, mu(11) = -1, mu(22) = 1.
• Sum over all divisors: 1 + (-1) + (-1) + 1 = 0.

The excerpt asks to compute this sum for other values and make a conjecture (the investigation guides the student to discover a pattern).

🔗 Multiplicative functions

🔗 Definition of multiplicativity

A function f on positive integers is multiplicative if f(N·M) = f(N)·f(M) whenever N and M are relatively prime (share no common prime factors).

• Relatively prime means gcd(N, M) = 1.
• The property is about factoring the function value, not about addition.

✅ All three functions are multiplicative

Function	Multiplicative?	Example showing failure when not relatively prime
sigma-zero	Yes	Must provide example where gcd(N,M) > 1
sigma-one	Yes	Must provide example where gcd(N,M) > 1
mu	Yes	Must provide example where gcd(N,M) > 1

• The excerpt asks students to verify multiplicativity and to find counterexamples when N and M share a common factor.
• Don't confuse: multiplicativity requires the inputs to be relatively prime; the property does not hold in general.

🧭 Overview

🧠 One-sentence thesis

The number of ways to choose k objects from n objects depends critically on whether order matters and whether repeats are allowed, yielding four distinct counting scenarios.

📌 Key points (3–5)

• The central question: "How many ways can you choose k objects from n objects?" has no single answer—it depends on two factors.
• Two key distinctions: whether order matters (does sequence count?) and whether repeats are allowed (can you pick the same object more than once?).
• Four different answers: the same problem (e.g., choosing 2 fruits from 5) gives 10, 20, 25, or 15 depending on the combination of order/repeat rules.
• Common confusion: "choosing 2 from 5" sounds like one problem, but you must always clarify order and repeats before counting.
• Goal of the chapter: find general formulas for each of the four cases when choosing k objects from n objects.

🍎 The four scenarios illustrated

🍎 Order does not matter, repeats not allowed

This is the classic "combination" scenario: selecting a subset where sequence is irrelevant and no duplicates.

• Example from the excerpt: Pick 2 pieces of fruit from 5 types (apple, banana, orange, pear, mango) to bring to work.
• You care only which fruits, not in what order you grab them.
• You cannot pick the same fruit twice (only one of each type in the fridge).
• Count: "5 choose 2" = 5 factorial divided by (2 factorial times 3 factorial) = 10 ways.

🗓️ Order matters, repeats not allowed

Now sequence is important, but you still cannot reuse an object.

• Example from the excerpt: Eat one fruit on Monday and a different fruit on Tuesday.
• Monday vs. Tuesday matters: apple-then-banana is different from banana-then-apple.
• Once you eat a fruit on Monday, it's gone—only 4 choices remain for Tuesday.
• Count: 5 choices for Monday × 4 choices for Tuesday = 20 ways.

🔁 Order matters, repeats allowed

Sequence matters and you can pick the same object multiple times.

• Example from the excerpt: The fridge is packed with plenty of each fruit type; you can eat an apple both Monday and Tuesday.
• Order still matters (Monday vs. Tuesday).
• Each day you have all 5 choices available again.
• Count: 5 choices for Monday × 5 choices for Tuesday = 25 ways.

🎒 Order does not matter, repeats allowed

You care only about the collection (not sequence), but duplicates are permitted.

• Example from the excerpt: Grab 2 pieces of fruit for your lunch bag; the fridge has plenty of each type.
• You can pick two of the same kind (e.g., two apples).
• Order doesn't matter: apple-apple is just one outcome, not two.
• Count: 5 ways to pick two of the same kind + 10 ways to pick two different kinds = 15 ways.
• Don't confuse: this is not the same as "order matters, repeats allowed" (25); allowing repeats when order doesn't matter requires careful counting to avoid double-counting pairs.

📊 Summary table

The excerpt provides a table that captures all four cases for choosing 2 fruits from 5 types:

	Order does not matter	Order matters
Repeats not allowed	10	20
Repeats allowed	15	25

• Each cell answers the same base question ("choose 2 from 5") under different rules.
• The chapter's goal is to generalize these numbers: find formulas for choosing k objects from n objects in each of the four cases.

🎯 What the chapter will cover

🎯 General formulas for all four cases

• The excerpt states that the chapter will study each case "in depth."
• The aim is to compute the number of ways to choose k objects from n objects for every combination of order/repeat rules.
• Throughout, k and n are assumed to be natural numbers (positive integers).

🎯 Why this matters

• The same counting question can have radically different answers depending on context.
• Before applying any formula, you must first determine:
  1. Does order matter in this situation?
  2. Are repeats allowed?
• Example: "How many ways can 7 people fill 4 student government positions?" depends on whether the positions are distinct roles (order matters) or just a committee (order does not matter).

🧭 Overview

🧠 One-sentence thesis

Sequences provide a framework for counting ordered arrangements of objects, with the formula n^k covering cases where order matters and repeats are allowed, and n! covering permutations where all n objects are arranged without repetition.

📌 Key points (3–5)

• What a sequence is: an ordered list of numbers or symbols, either finite or infinite, written with parentheses and commas like (5, 2, 3, 6).
• Counting sequences with repeats allowed: when choosing k objects from n objects where order matters and repeats are allowed, there are n^k possibilities.
• Permutations (no repeats): a permutation arranges all n objects in order with no repeats; there are n! permutations of n objects.
• Common confusion: the fruit-choosing examples show that "choose 2 from 5" can yield 10, 20, 25, or 15 depending on whether order matters and whether repeats are allowed—context determines the correct formula.
• Generalization with different alphabets: when each position in a sequence can draw from a different set of symbols (sizes n₁, n₂, ..., nₖ), the total count is n₁ · n₂ · ... · nₖ.

📝 What sequences are

📝 Definition and notation

Sequence: a list of numbers or other symbols written in order; can be finite or infinite.

• A finite sequence has a specific length (number of entries), e.g., (5, 2, 3, 6) has length 4.
• An infinite sequence continues indefinitely, e.g., (2, 4, 6, 8, 10, ...).
• Notation: (a₁, a₂, ..., aₙ) denotes a finite sequence of length n; (a₁, a₂, a₃, ...) denotes an infinite sequence where the i-th entry is aᵢ for each integer i ≥ 1.

🔤 Alphabet and strings

Alphabet: the set of numbers or symbols used in a sequence.

• Common alphabets: binary {0, 1}, English letters, ASCII, or natural numbers.
• String or word: a sequence written without parentheses and commas for convenience, e.g., (a, p, p, l, e) becomes "apple" and (1, 0, 1, 1, 0) becomes "10110".
• Warning: don't write (1, 0, 10, 100, 1000) as "10101001000" because you lose information about entry boundaries.

🔢 Alternative view as functions

• A sequence of length n can be thought of as a function assigning a number or symbol to every positive integer from 1 to n.
• An infinite sequence assigns a number or symbol to every positive integer.

🔄 Zero-indexing variation

• Sometimes it is convenient to start a sequence with a₀ rather than a₁ (zero-indexing).

🧮 Counting sequences: order matters, repeats allowed

🧮 The n^k formula

Theorem: The number of sequences of length k composed from an alphabet with n symbols is n^k.

• More generally: there are n^k combinations of k objects chosen from n objects where order matters and repeats are allowed.
• Why: there are n ways to choose the first symbol, n ways to choose the second, and so on up to the k-th, giving n · n · n · ... · n = n^k possibilities.
• This fills one entry in the "choose k from n" table:

	Order does not matter	Order matters
Repeats not allowed	?	?
Repeats allowed	?	n^k

🔢 Examples with uniform alphabets

Binary strings of length k:

• There are 2^k binary strings (n = 2).
• Example: for k = 3, all 8 = 2³ strings are 000, 001, 010, 011, 100, 101, 110, 111.

Three-letter strings from English alphabet:

• There are 26³ = 17,576 possibilities, e.g., "pxy" or "csu".

Three-character strings (letters or digits):

• There are (26 + 10)³ = 36³ = 46,656 possibilities, e.g., "p17", "2xy", or "pqm".

Buffet snacks:

• With 3 types of snacks (samosas, chicken skewers, carrots) and eating 10 snacks in a row, there are 3¹⁰ ways.

🔀 Generalization: different alphabets per position

Theorem: There are n₁ · n₂ · ... · nₖ possible sequences of length k made by choosing one of n₁ symbols as the 1st entry, one of n₂ symbols as the 2nd entry, ..., one of nₖ symbols as the k-th entry.

• The n^k formula is the special case where n₁ = n₂ = ... = nₖ = n.

Example: 4-digit numbers with no leading zeros:

• First digit: 9 choices (1–9).
• Remaining three digits: 10 choices each (0–9).
• Total: 9 · 10 · 10 · 10 = 9,000.

Example: consonant-vowel-consonant strings:

• First letter: 21 consonants.
• Middle letter: 5 vowels (A, E, I, O, U).
• Last letter: 21 consonants.
• Total: 21 · 5 · 21 = 2,205.

🔄 Permutations: order matters, no repeats

🔄 What a permutation is

Permutation: an ordering or arrangement of all n objects; can be expressed as a sequence or string.

• A permutation uses each object exactly once.
• Example: the set {a, b, c} has 6 permutations: abc, acb, bac, bca, cab, cba.
• For sets with multi-digit numbers like {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, use sequence notation to avoid confusion, e.g., (3, 6, 7, 2, 1, 10, 4, 5, 9, 8).

🔄 The n! formula

Theorem: The number of permutations of n objects is n! (n factorial), where n! = n · (n − 1) · (n − 2) · ... · 3 · 2 · 1.

• Why: there are n choices for the first object, n − 1 for the second, n − 2 for the third, and so on until only 1 choice remains for the n-th object, giving n · (n − 1) · (n − 2) · ... · 1 = n!.
• Special case: 0! = 1, because there is one way to arrange zero objects (the "empty sequence" or "empty string").

🍦 Examples of permutations

Seven students lining up:

• There are 7! = 5,040 ways for seven students {a, b, c, d, e, f, g} to line up for ice cream.

Five fruits for five weekdays:

• Assign one of 5 different fruits (apple, banana, orange, pear, mango) to each day of the work week, using each fruit exactly once.
• This is a permutation of {A, B, O, P, M}.
• Total: 5! = 5 · 4 · 3 · 2 · 1 = 120.

🔍 Don't confuse permutations with sequences allowing repeats

• Permutations require all n objects used exactly once (no repeats, all objects included).
• Sequences with repeats allowed (n^k) let you reuse objects and may not use all objects.
• Example: choosing 2 fruits from 5 where order matters gives 5 · 4 = 20 if repeats are not allowed (a partial permutation, covered in the next section) but 5² = 25 if repeats are allowed.

🔗 Preview: ordered subsets

🔗 What comes next

• The excerpt introduces the concept of ordered subsets: choosing k objects from n objects where order matters and repeats are not allowed.
• This is a generalization of permutations (which arrange all n objects) to cases where only k < n objects are selected and arranged.
• The excerpt ends before providing the formula or detailed examples for ordered subsets, noting that the ice cream adventurers face a new problem requiring this concept.

🧭 Overview

🧠 One-sentence thesis

The number of ways to arrange n objects in order is n factorial, and when arranging only k objects from n, the count is n!/(n−k)!, both representing scenarios where order matters and repeats are not allowed.

📌 Key points (3–5)

• Permutations count all arrangements: A permutation arranges all n objects in order, and there are n! such arrangements.
• Ordered subsets for partial arrangements: When choosing and ordering only k objects from n (where k ≤ n), there are n!/(n−k)! ways.
• Order matters, no repeats: Both permutations and ordered subsets require distinct entries—no object appears twice in the same arrangement.
• Common confusion: Ordered subsets vs permutations—ordered subsets arrange only k out of n objects, while permutations arrange all n; when k = n, the formulas coincide.
• Building block for the counting table: These formulas fill the "order matters, repeats not allowed" row of the general counting framework.

🔢 Permutations: arranging all objects

🔢 What a permutation is

Permutation of n objects: an ordering or arrangement of all n of them, expressed as a sequence or string.

• A permutation uses every object exactly once.
• The order in which objects appear matters—different orders are different permutations.
• Example: The set {a, b, c} has 6 permutations: abc, acb, bac, bca, cab, cba.

🧮 Counting permutations: n! formula

Theorem: The number of permutations of n objects is n!.

• Why n!: By the multiplication principle, there are n choices for the first position, then n−1 for the second, n−2 for the third, and so on down to 1 choice for the last position.
• The product n · (n−1) · (n−2) · ... · 3 · 2 · 1 equals n!.
• Example: Seven students lining up for ice cream can arrange themselves in 7! = 5,040 ways.

🍎 Everyday scenarios

• Fruit assignment: If you have 5 different fruits and want to bring one each day of a 5-day work week (no repeats), there are 5! = 120 ways to assign them.
• License plates and codes: The excerpt mentions problems like counting license plate combinations and donor account codes, which use similar multiplication principles.

🕳️ The empty case: 0! = 1

• When there are 0 objects, there is exactly one way to arrange them: do nothing (the "empty sequence").
• By convention, 0! = 1 to make formulas consistent.

📦 Ordered subsets: arranging k out of n objects

📦 What an ordered subset is

Ordered subset of size k from a set A: a string of k distinct elements chosen from A, where order matters.

• Unlike a full permutation, an ordered subset arranges only k objects (k ≤ n), not all n.
• Each element in the string must be distinct—no repeats.
• Example: Choosing 4 students from 7 to form a line (where the shop can only serve 4 customers) creates an ordered subset of size 4.

🧮 Counting ordered subsets: n!/(n−k)!

Theorem: The number of ordered subsets of size k from a set with n elements is
n · (n−1) · (n−2) · ... · (n−k+1) = n! / (n−k)!.

• Why this formula:
  • There are n choices for the first position.
  • Then n−1 for the second, n−2 for the third, and so on.
  • For the kth (final) position, there are n−k+1 remaining choices.
  • The product has exactly k factors.
• The decreasing number of choices at each step ensures no repeated entries.
• Example: Forming a line of 4 students from 7 gives 7 · 6 · 5 · 4 = 840 ways, which equals 7! / 3!.

🍊 Practical examples

Scenario	n	k	Count	Explanation
Fruits on Monday and Tuesday	5	2	5!/3! = 20	Choose and order 2 fruits from 5
47 students in 50 seats	50	47	50!/3!	Assign 47 students to 47 of 50 seats in order
Visit 5 state capitals from 50	50	5	50!/45!	Order matters: Sacramento→Dover→... differs from Dover→...

💡 Computation tip

• When dividing factorials, expand both and cancel before multiplying.
• Example: 5!/3! = (5·4·3·2·1)/(3·2·1) = 5·4 = 20, which is easier than computing 120/6.

🔄 Don't confuse: ordered subsets vs permutations

• Permutations arrange all n objects (k = n).
• Ordered subsets arrange only k objects from n (k can be less than n).
• When k = n, the ordered subset formula n!/(n−k)! = n!/0! = n! reduces to the permutation formula.

🗂️ Filling the counting table

🗂️ The "order matters, repeats not allowed" row

The excerpt completes part of a general counting framework:

Condition	Order doesn't matter	Order matters
Repeats not allowed	(n choose k) = n!/(k!(n−k)!)	n!/(n−k)!
Repeats allowed	(formula not in this excerpt)	n^k

• The right column's top entry, n!/(n−k)!, counts ordered subsets (this section's focus).
• The left column's top entry, (n choose k), counts unordered subsets (covered in Section 3.4, briefly mentioned here).
• The bottom-right entry, n^k, was covered earlier (repeats allowed, order matters).

🔗 Connection to unordered subsets (preview)

• The excerpt mentions that (n choose k) = n!/(k!(n−k)!) counts subsets where order does not matter.
• The proof idea: ordered subsets count each unordered subset k! times (once for each permutation of the k chosen objects), so divide n!/(n−k)! by k! to get (n choose k).
• Example: The ice cream contest chooses 4 winners from 107 entrants; since all winners receive the same prize, order doesn't matter, so the answer is (107 choose 4).

🧩 Key distinctions and reminders

🧩 Order matters vs order doesn't matter

• Order matters: "First place, second place, third place" are different roles → use n!/(n−k)!.
• Order doesn't matter: "Four winners all get the same prize" → use (n choose k).
• Example: Ranking your top 10 restaurants (#1 to #10) from 50 uses ordered subsets (order matters); choosing 4 contest winners from 107 uses unordered subsets (order doesn't matter).

🧩 Repeats allowed vs repeats not allowed

• Repeats not allowed: Each object used at most once → formulas involve factorials (n! or n!/(n−k)!).
• Repeats allowed: Objects can be reused → formula is n^k (from earlier sections).
• Example: Scrambling 7 distinct letters into a 7-letter word (each letter used exactly once) is a permutation (7!); scrambling into a 5-letter word (each letter used at most once) is an ordered subset (7!/2!).

🧩 Practical problem-solving checklist

1. Does order matter? (Line order, ranking, sequence → yes; prize winners, committee members → no)
2. Are repeats allowed? (Distinct entries required → no; can reuse → yes)
3. Are we arranging all n or only k? (All n → permutation n!; only k → ordered subset n!/(n−k)!)

Example: "How many ways to stack 5 different rocks with your favorite on top?" → Order matters (stack position), no repeats (distinct rocks), arrange all 5 but one position is fixed → 4! ways to arrange the remaining 4 rocks below the top one.

🧭 Overview

🧠 One-sentence thesis

The binomial coefficient formula counts how many ways you can choose k objects from n objects when order does not matter and repeats are not allowed, and this counting principle extends to surprising applications like binary strings and multisets.

📌 Key points (3–5)

• What the binomial coefficient counts: the number of subsets of size k from a set of size n, where order does not matter and repeats are not allowed.
• How it relates to ordered counting: the formula divides ordered subsets by k! because each unordered subset is counted k! times when order matters.
• Common confusion: choosing k winners is the same as eliminating (n − k) losers, so (n choose k) equals (n choose n − k).
• Surprising application: the formula also counts binary strings with exactly k zeroes and (n − k) ones, even though strings are ordered.
• Why it matters: completes the counting table for "order does not matter, repeats not allowed" and provides a foundation for multiset counting.

🧮 The binomial coefficient formula

🧮 What it counts

The number of subsets of size k in a set of size n is (n choose k) = n! / (k!(n − k)!).

• More generally, this is the number of ways to choose k objects from n objects where order does not matter and repeats are not allowed.
• This fills in the upper-left entry of the counting table.

Situation	Order does not matter	Order matters
Repeats not allowed	(n choose k)	n! / (n − k)!
Repeats allowed	(to be covered)	n^k

🔍 Why the formula works (proof sketch)

The proof uses the division principle:

1. Start with ordered subsets: by an earlier theorem, the number of ordered subsets of size k is n! / (n − k)!.
2. Each unordered subset is overcounted: every unordered subset gets counted k! times (once for each permutation of the k objects).
3. Divide to correct the overcount: divide n! / (n − k)! by k! to get n! / (k!(n − k)!).

• Example: If you have 5 fruits and choose 2 to bring to work on the same day (order doesn't matter), the answer is (5 choose 2) = 5! / (2! · 3!) = (5 · 4 · 3 · 2 · 1) / (2 · 1 · 3 · 2 · 1) = 5 · 2 = 10.

💡 Computation tip

When dividing one factorial by another, expand both as products and cancel terms before multiplying.

• Don't compute 5! = 120 and 3! = 6 separately and then divide.
• Instead, write 5! / 3! = (5 · 4 · 3 · 2 · 1) / (3 · 2 · 1), cancel the 3 · 2 · 1, and get 5 · 4 = 20 directly.

🎯 Worked examples

🎯 Ice cream contest winners

Problem: Out of 107 entrants, how many ways are there to choose 4 winners for an "Ice Cream for a Year" contest? All winners receive the same prize.

Answer: (107 choose 4) = 107! / (4! · 103!) = 5,160,610.

• Order does not matter because all winners get the same prize.
• Repeats are not allowed because each person can win at most once.

🏈 NCAA football teams (unranked)

Problem: How many ways are there to select the top 25 NCAA DI football teams (out of 128) without ranking them?

Answer: (128 choose 25) = 128! / (25! · 103!).

• This is the number of unordered subsets of size 25.
• Don't confuse with the earlier exercise about ranking: if you rank them, order matters and the answer is 128! / 103!.

🎁 Wrapping teddy bears

Problem: You have 7 different rolls of wrapping paper and 4 identical teddy bears. How many ways can you wrap the teddy bears such that no two bears are wrapped in the same paper?

Answer: (7 choose 4).

• You are choosing 4 different wrapping papers out of 7.
• Order does not matter because the teddy bears are identical.

🔄 Symmetry property

🔄 Choosing vs eliminating

Key insight: Choosing k winners out of n entrants is the same as eliminating (n − k) losers.

• In the ice cream contest, choosing 4 winners out of 107 is the same as eliminating 103 people who did not win.
• Both give the same count: (107 choose 4) = (107 choose 103).

📐 Corollary formula

(n choose k) = (n choose n − k).

• This symmetry often simplifies calculations.
• Example: (128 choose 25) = (128 choose 103), so you can compute whichever is easier.

🔢 Surprising application: binary strings

🔢 Counting binary strings with fixed numbers of 0s and 1s

Problem: How many binary strings (from the alphabet {0, 1}) have exactly k zeroes and (n − k) ones?

Answer: (n choose k).

Why this works:

• A binary string of length n has n positions.
• To form a string with exactly k zeroes, you simply choose k positions (out of n) for the 0's.
• The remaining (n − k) positions automatically get 1's.
• Choosing k positions is an unordered choice, even though the string itself is ordered.

Don't confuse: The string is ordered (position matters), but the choice of which positions get 0's is unordered (you are choosing a subset of positions).

• Example: If n = 5 and k = 2, there are (5 choose 2) = 10 binary strings with exactly two 0's and three 1's.

🧩 Transition to multisets

🧩 What's next: repeats allowed

The excerpt introduces the next topic: multisets, which are sets where elements can repeat (have multiplicity greater than 1).

• A multiset is a set together with a positive integer multiplicity assigned to each element.
• Example: {apple, apple, orange, pear, pear} is a multiset with two apples, one orange, and two pears.
• Notation: The number of multisets of size k from n types is written ((n multichoose k)) or ((n choose k with repeats)).

🍦 Ice cream sundae problem

Setup: There are 10 flavors of ice cream and a large sundae has 15 scoops. Order does not matter (the scoops melt together) and repeats are allowed (you can have multiple scoops of the same flavor). How many different sundaes can you order?

• This is a "repeats allowed, order does not matter" problem.
• The excerpt begins to introduce the "sticks and stones" method (also called "stars and bars") to count multisets, but the full solution is not provided in this excerpt.

🧭 Overview

🧠 One-sentence thesis

Multisets extend the idea of sets by allowing repeated elements, and counting them uses the "sticks and stones" method to transform the problem into choosing positions for dividers and items.

📌 Key points (3–5)

• What a multiset is: a collection where the same element can appear multiple times, each with a specified multiplicity.
• The counting formula: the number of multisets of size k from n types equals (n+k−1 choose k), which is written as ((n choose k with repeats)).
• Sticks and stones method: represent multisets as strings of k stones (items) and n−1 sticks (dividers between types), then count positions.
• Common confusion: don't confuse multisets with ordinary sets—order still doesn't matter, but repetition is now allowed.
• Applications: multisets solve distribution problems (identical objects to people) and composition problems (ordered lists of nonnegative integers with fixed sum).

🎯 What multisets are

🎯 Definition and notation

Multiset: a set together with a positive integer multiplicity assigned to each element; equivalently, a set S with a function f : S → N − {0}.

• Unlike ordinary sets, elements can repeat.
• Example: {apple, apple, orange, pear, pear} is a multiset with two apples and two pears.

✍️ Ways to write multisets

Three equivalent notations:

1. List with repeats: {4, 4, 4, 5, 6, 6} or {4, 5, 4, 6, 4, 6} (order doesn't matter)
2. Multiplicity notation: {4×3, 5×1, 6×2} (the ×3 means "occurs 3 times," not exponentiation)
3. Function form: f : {4, 5, 6} → N with f(4) = 3, f(5) = 1, f(6) = 2

🔢 Notation for counting

• ((n choose k)) or ((n multichoose k)): the number of multisets of size k whose elements come from n types.
• Pronounced "n multichoose k" or "n choose k with repeats."

🪨 The sticks and stones counting method

🪨 How the method works

The key insight: represent each multiset as a string of stones and sticks.

• Stones (•): represent individual items
• Sticks (|): act as dividers between different types

Example: A bag with 1 apple, 2 bananas, 2 mangoes, 0 oranges, 1 pear (from 5 fruit types A, B, M, O, P) becomes:

• Multiset: {A, B, B, M, M, P}
• String: • | • • | • • | | •

The empty space between the last two sticks shows zero oranges.

🧮 Why it gives the formula

For k items from n types:

• Need k stones (the items)
• Need n−1 sticks (to separate n types)
• Total positions: k + (n−1) = n + k − 1
• Choose which k positions get stones: (n+k−1 choose k)

Theorem: The number of multisets of size k from n types equals (n+k−1 choose k).

🍦 Ice cream example

Student c orders a 15-scoop sundae from 10 flavors:

• k = 15 scoops, n = 10 flavors
• Answer: (10+15−1 choose 15) = (24 choose 15) = 1,307,504 different sundaes
• That's about 3,593 sundaes per day to try them all in a year!

🎈 Balloon example

A store sells 6 colors of balloons; you buy 10 balloons:

• k = 10 balloons, n = 6 colors
• Answer: ((6 choose 10)) = (6+10−1 choose 10) = (15 choose 10) = 3,003 ways

🎁 Variation: at least one of each type

🎁 The constraint changes the count

Problem: Choose m elements from n types where each type must appear at least once (m ≥ n).

Solution approach:

1. First, pick one of each type (uses n items)
2. Then choose the remaining m − n items freely from n types
3. Apply the multiset formula to the remaining items

Corollary: The number of ways is (m−1 choose n−1).

🍦 Sundae with every flavor

If the sundae must have at least one scoop of each of the 10 flavors:

• First 10 scoops determined (one per flavor)
• Choose 5 more scoops from 10 flavors
• Answer: (14 choose 5) = 2,002 sundaes (about 6 per day for a year)

🎈 Bouquet with every color

10 balloons from 6 colors, at least one of each color:

• m = 10, n = 6
• Answer: (10−1 choose 6−1) = (9 choose 5)

💰 Distribution problems

💰 Distributing identical objects

Corollary: The number of ways to distribute k identical pennies to n people is (n+k−1 choose k).

Why it works:

• Label people A₁, A₂, ..., Aₙ
• Create a multiset where each person's label appears as many times as pennies they receive
• This is exactly a multiset of size k from n types

Example: 5 pennies to 4 people

• Multiset {A₁, A₁, A₃, A₄, A₄} means: person A₁ gets 2, A₂ gets 0, A₃ gets 1, A₄ gets 2
• Total ways: (4+5−1 choose 5) = (8 choose 5) = 56

💰 Distribution with minimum requirement

If each person must receive at least one penny:

• First give 1 penny to each person (uses n pennies)
• Distribute remaining m − n pennies freely
• Answer: (m−1 choose n−1)

Example: 9 pennies to 4 people, each gets at least one

• Answer: (9−1 choose 4−1) = (8 choose 3) = 56

📝 Compositions: ordered lists with fixed sum

📝 What compositions count

Composition: an ordered list of nonnegative integers with fixed length and sum.

Example: Compositions of 9 of length 4 include:

• (4, 0, 2, 3)
• (0, 4, 2, 3)
• (0, 0, 0, 9)

Note: (4, 0, 2, 3) and (0, 4, 2, 3) are different because order matters.

📝 Connection to distribution

A composition of m of length n encodes distributing m identical pennies to n people:

• Position i in the list = number of pennies person i receives
• Therefore: number of compositions of m of length n = (n+m−1 choose m)

Example: Compositions of 9 of length 4

• Same as distributing 9 pennies to 4 people
• Answer: ((4 choose 9)) = (12 choose 9) = 220

📝 Compositions with positive integers

If all entries must be positive (at least 1):

• Each person must get at least one penny
• Subtract 1 from each position
• Reduces to compositions of m−n of length n
• Answer: (m−1 choose n−1)

Example: Ordered lists of 4 positive integers summing to 9

• Same as compositions of 5 of length 4
• Answer: (8 choose 3) = 56

📊 Summary table

📊 Complete counting formulas

Scenario	Order doesn't matter	Order matters
Repeats not allowed	(n choose k)	n! / (n−k)!
Repeats allowed	(n+k−1 choose k)	nᵏ

The multiset formula (n+k−1 choose k) completes the lower-left cell of this fundamental counting table.

🧭 Overview

🧠 One-sentence thesis

This section consolidates the four major counting formulas that distinguish whether order matters and whether repetition is allowed when choosing k objects from n objects.

📌 Key points (3–5)

• What the section provides: a summary table of the four fundamental counting formulas from the chapter.
• Two key dimensions: order (does it matter or not?) and repetition (allowed or not allowed?).
• Common confusion: distinguishing when to use combinations vs permutations, and when repetition changes the formula.
• The four cases: no-order/no-repeat uses binomial coefficients; order/no-repeat uses factorial division; no-order/with-repeat uses stars-and-bars; order/with-repeat uses powers.

📊 The four counting formulas

📊 Summary table

The excerpt presents a table organizing the formulas by two criteria:

Order matters?	Repeats allowed?	Formula	Notation
No	No	n choose k	(n choose k)
No	Yes	n + k − 1 choose k	((n + k − 1) choose k)
Yes	No	n factorial divided by (n − k) factorial	n! / (n − k)!
Yes	Yes	n to the power k	n^k

🔍 What each formula counts

• (n choose k): choosing k objects from n when order does not matter and you cannot pick the same object twice.

  • Example: selecting 3 students from a class of 10 for a committee.

• ((n + k − 1) choose k): choosing k objects from n types when order does not matter but you can pick the same type multiple times.

  • Example: distributing 10 identical candies to 4 students (stars-and-bars).

• n! / (n − k)!: arranging k objects chosen from n when order matters and no repetition is allowed.

  • Example: assigning 1st, 2nd, 3rd place from 8 contestants.

• n^k: arranging k objects from n types when order matters and repetition is allowed.

  • Example: creating a 5-character password where each character can be any of 26 letters.

🧩 How to choose the right formula

🧩 Does order matter?

• Order matters means different arrangements of the same objects count as different outcomes.

  • If you care about sequence or position → use the "order matters" row.

• Order does not matter means you only care about which objects are selected, not their arrangement.

  • If you care only about the set of chosen objects → use the "order does not matter" row.

🔁 Are repeats allowed?

• Repeats not allowed means each object can be chosen at most once.

  • Typical in selection problems where objects are distinct and cannot be reused.

• Repeats allowed means the same object (or type) can be chosen multiple times.

  • Common in distribution problems with identical items or when sampling with replacement.

⚠️ Common confusion: combinations with repetition

• Don't confuse the "no order, repeats allowed" case with the basic binomial coefficient.
• The formula ((n + k − 1) choose k) accounts for distributing k identical items into n bins (stars-and-bars method).
• Example: How many ways to buy 10 pieces of sushi from 3 types? Use ((3 + 10 − 1) choose 10) = ((12) choose 10), not (3 choose 10).

🎯 Context and application

🎯 What this summary consolidates

• The excerpt states these are "the major results in this chapter."
• The formulas are presented without derivation; they summarize earlier sections (likely Sections 3.1–3.5).
• The table is a reference tool for quickly identifying which formula applies to a given counting problem.

🎯 How to use the table

1. Identify whether the problem cares about order (arrangement vs selection).
2. Determine whether the same object can appear more than once.
3. Locate the corresponding cell in the table.
4. Apply the formula with the appropriate values of n (number of object types or total objects) and k (number of choices or positions).

🧭 Overview

🧠 One-sentence thesis

This section provides practice problems that apply the counting formulas from Chapter 3 to a variety of scenarios including permutations, combinations with and without repetition, and both ordered and unordered selections.

📌 Key points (3–5)

• Core formulas to apply: the chapter summary table shows four counting formulas based on whether order matters and whether repeats are allowed.
• Problem types: permutations of letters with repetition, distributing identical or distinct objects, forming strings with constraints, and counting outcomes with dice or cards.
• Common confusion: distinguishing when order matters vs. when it does not—strings and ranked lists are ordered; distributing identical items to people (without ranking the people) is unordered.
• Repeats allowed vs. not allowed: giving 20 identical rocks to 12 students allows repeats (one student can get multiple rocks); choosing distinct letters does not.
• Real-world application: the poker investigation (section 3.8) demonstrates how to count card hands by breaking down constraints (suits, values, runs, flushes).

📊 Summary of counting formulas

📊 The four core formulas

The excerpt provides a table summarizing the major results:

Order matters?	Repeats allowed?	Formula	Meaning
No	No	(n choose k)	Choose k objects from n, no order, no repeats
Yes	No	n! / (n - k)!	Arrange k objects from n, order matters, no repeats
No	Yes	(n + k - 1 choose k)	Choose k objects from n, no order, repeats allowed
Yes	Yes	n^k	Arrange k objects from n, order matters, repeats allowed

• These formulas are the backbone for solving all the problems in section 3.7.
• The table helps you decide which formula to use based on the problem's constraints.

🔤 Permutation and arrangement problems

🔤 Permutations with repeated letters

Problem 1: How many different strings are permutations of the letters in PILLOW?

• The word PILLOW has repeated letters (two L's).
• This is a permutation problem where order matters but some objects are identical.
• Don't confuse: if all letters were distinct, the answer would be 6 factorial; repetition reduces the count because swapping identical letters does not create a new permutation.

🏆 Ranked lists and ordered selections

Problem 2: Baseball team A has 7 pitchers, team B has 5 catchers. A trade involves 3 of A's pitchers for 2 of B's catchers. The newspaper makes a ranked list of their top 4 favorite possible trades. How many different ranked lists can the newspaper make?

• First, count the number of possible trades: choose 3 pitchers from 7 and 2 catchers from 5 (order does not matter within a trade).
• Then, arrange 4 trades into a ranked list (order matters for the ranking).
• Example: Trade X ranked first is different from Trade X ranked second.

🔢 Strings with mixed constraints

Problem 3: How many ordered strings with 10 symbols have exactly 7 numbers (0–9) and exactly 3 letters (a–z)?

• Order matters: the string abc1222233 is different from 12c2a3b322.
• Step 1: choose 7 positions out of 10 for the numbers (the remaining 3 are for letters).
• Step 2: fill the 7 number positions (repeats allowed, 10 choices each).
• Step 3: fill the 3 letter positions (repeats allowed, 26 choices each).

🪨 Distribution problems

🪨 Distributing identical objects

Problem 4(a): Find the number of ways to give 20 identical rocks to 12 students (a through l).

Distributing identical objects to distinct recipients with no restrictions: use the formula (n + k - 1 choose k), where n is the number of recipients and k is the number of objects.

• Here, order does not matter (we don't rank the students), and repeats are allowed (one student can receive multiple rocks).
• This is the "stars and bars" problem.

🪨 Distributing with a minimum constraint

Problem 4(b): Same as (a), but every student receives at least one rock.

• First, give 1 rock to each of the 12 students (using 12 rocks).
• Then, distribute the remaining 20 - 12 = 8 rocks with no restrictions.
• Don't confuse: "at least one" changes the problem by reducing the number of objects left to distribute freely.

🪨 Distributing distinct objects

Problem 4(c): The rocks are all different and there are no restrictions (one student can receive all 20 rocks).

• Each of the 20 distinct rocks can go to any of the 12 students.
• Order matters in the sense that each rock's assignment is independent.
• This is the formula n^k (12 choices for each of 20 rocks).

🧬 String and sequence problems

🧬 Sequences with a fixed alphabet

Problem 5: How many different DNA sequences of length 5 are there? (Alphabet: A, C, G, T)

• Order matters, repeats allowed.
• 4 choices for each of 5 positions → 4^5.

Problem 6: How many strings of four distinct letters from the English alphabet (26 letters)?

• Order matters, repeats not allowed (distinct means no letter appears twice).
• 26 choices for the first position, 25 for the second, 24 for the third, 23 for the fourth.
• Formula: 26! / (26 - 4)!

🧬 Binary strings with balance constraints

Problem 7: How many binary strings of length 8 have the same number of 0's as 1's?

• Length 8 with equal 0's and 1's means exactly 4 zeros and 4 ones.
• Choose 4 positions out of 8 for the zeros (the rest are ones).
• Formula: (8 choose 4).

🧬 Strings with exact or bounded counts

Problem 8: How many strings of length 7 from the alphabet {0, 1, 2} have exactly three 0's?

• Choose 3 positions out of 7 for the 0's.
• Fill the remaining 4 positions with 1 or 2 (2 choices each).
• Formula: (7 choose 3) × 2^4.

Problem 9: How many strings of length 7 from {0, 1, 2} have less than three 0's?

• "Less than three" means 0, 1, or 2 zeros.
• Count each case separately and add them up.
• Don't confuse: "less than three" is not the same as "exactly three."

🎲 Dice and card problems

🎲 Rolling dice with a sum constraint

Problem 10: How many ways can you roll a sum of 13 with three six-sided dice?

• Each die shows 1 through 6.
• Order matters (die 1, die 2, die 3 are distinguishable).
• Count all ordered triples (a, b, c) where a + b + c = 13 and 1 ≤ a, b, c ≤ 6.

🃏 Poker hand counting (section 3.8 preview)

The excerpt introduces a poker investigation that applies counting formulas to card hands.

Key poker vocabulary:

• A standard deck has 52 cards: 4 suits (hearts, spades, diamonds, clubs), 13 values per suit (2–10, Jack, Queen, King, Ace).
• A hand is a set of 5 cards.
• A run is 3 or more consecutive values (suit does not matter); Ace can be low or high but not in the middle.

Example hands and counts:

Hand type	Definition	Example count formula	Result
Royal flush	Run of A, K, Q, J, 10, all same suit	(4 choose 1)	4
Straight flush	Run of 5, all same suit (not royal)	(10 choose 1) × (4 choose 1) - 4	36
Four of a kind	Four cards of one value, any fifth card	(13 choose 1) × (48 choose 1)	624
Full house	Three of one value, two of another	(13 choose 1) × (4 choose 3) × (12 choose 1) × (4 choose 2)	(formula given, result cut off in excerpt)

• Royal flush: values are fixed, so only choose the suit.
• Straight flush: choose the starting value (10 options: Ace through 10) and the suit, then subtract the 4 royal flushes.
• Four of a kind: choose the value for the four cards, then choose any of the remaining 48 cards for the fifth.
• Full house: choose the value for the three cards, pick 3 suits out of 4 for them, choose a different value for the pair, pick 2 suits out of 4 for the pair.

Don't confuse: a run (straight) cares about consecutive values but not suit; a flush cares about all cards being the same suit but not consecutive values; a straight flush requires both.

🧭 Overview

🧠 One-sentence thesis

Counting poker hands requires systematically choosing card values and suits while carefully subtracting overlapping cases to avoid double-counting hands that belong to higher-ranking categories.

📌 Key points (3–5)

• What we're counting: the number of distinct five-card hands that satisfy specific poker hand definitions, using combinations and the multiplication principle.
• Core method: choose values first, then suits, then subtract any hands that would belong to a better (already-counted) category.
• Common confusion: order matters vs. order doesn't matter—when choosing the last few cards in a hand, dividing by factorial accounts for the fact that drawing cards in different orders produces the same hand.
• Why subtraction is necessary: some counting methods initially include hands that belong to higher-ranking categories (e.g., counting flushes initially includes straight flushes), so we must subtract those to get the correct count.
• Key constraint: a standard deck has 52 cards in 4 suits, with 13 values per suit; each hand contains exactly 5 cards.

🃏 Deck and poker basics

🃏 Standard deck structure

A standard deck contains 52 cards sorted into four suits (♥ hearts, ♠ spades, ♦ diamonds, ♣ clubs), with 13 cards in each suit.

• Each suit contains cards with values: 2, 3, 4, 5, 6, 7, 8, 9, 10, J (Jack), Q (Queen), K (King), A (Ace).
• A hand is a set of five cards held by a player.
• The excerpt emphasizes that 13 × 4 = 52 (quick check).

🔢 Runs and Aces

A run is a set of three or more cards with consecutive values where suit does not matter.

• Example: 3♥, 4♠, 5♠ is a run of three cards; 9♥, 10♠, J♣, Q♣, K♦ is a run of five.
• Special Ace rule: An Ace can be either the lowest or highest card in a run, but not in the middle.
  • Ace high: Q♣, K♦, A♦
  • Ace low: A♣, 2♥, 3♥
  • Not a run: K♥, A♦, 2♣ (Ace cannot bridge King and 2)

🏆 High-ranking hands (rare)

👑 Royal flush

A run consisting of an Ace, King, Queen, Jack, and 10, all of the same suit.

• Number of hands: (4 choose 1) = 4
• Why: The five card values are fixed (A, K, Q, J, 10), so we only choose one of the four suits.

🌊 Straight flush

A run of five cards, all of the same suit (but not a royal flush).

• Number of hands: (10 choose 1) × (4 choose 1) − 4 = 36
• Why:
  • Choose the starting card value: 10 ways (Ace through 10).
  • Choose the suit: 4 ways.
  • Subtract the 4 royal flushes (runs starting with 10).
• Don't confuse: If we tried to start with a Jack, we wouldn't have enough cards (Aces end runs).

🎯 Four of a kind

Four cards of the same value and any other card.

• Number of hands: (13 choose 1) × (48 choose 1) = 624
• Why:
  • Pick one value for the four cards: 13 ways.
  • Choose the fifth card from the remaining 52 − 4 = 48 cards.

🏠 Full house

Three cards of one value and two cards of a second value.

• Number of hands: (13 choose 1) × (4 choose 3) × (12 choose 1) × (4 choose 2) = 3,744
• Why:
  • Choose the value for the triple: 13 ways.
  • Choose 3 suits for those 3 cards: (4 choose 3) ways.
  • Choose the value for the pair from the remaining 12 values: 12 ways.
  • Choose 2 suits for those 2 cards: (4 choose 2) ways.

💧 Flush and straight hands

💧 Flush

All five cards have the same suit, but the hand is not a run (not a straight or royal flush).

• Number of hands: (4 choose 1) × (13 choose 5) − 40 = 5,108
• Why:
  • Choose one suit: 4 ways.
  • Choose 5 values from the 13 possible: (13 choose 5) ways.
  • Subtract 40 (the total number of straight flushes and royal flushes: 36 + 4).
• Key idea: The initial count includes some straight/royal flushes, so we must subtract them.

📏 Straight

A run of five cards with at least two suits (not all the same suit).

• Number of hands: (10 choose 1) × (4 choose 1)^5 − 40 = 10,200
• Why:
  • Pick the starting card value: 10 ways.
  • Pick a suit for each of the 5 cards: 4^5 ways.
  • Subtract 40 (straight flushes and royal flushes, which are all the same suit).

🎲 Mid-ranking hands (pairs and triples)

🎲 Three of a kind

Exactly three cards of the same value and two other cards with distinct values (not a full house).

• Number of hands: (13 choose 1) × (4 choose 3) × [(48 choose 1) × (44 choose 1)] / 2! = 54,912
• Why:
  • Choose the value and three suits for the triple: (13 choose 1) × (4 choose 3).
  • Choose the first remaining card from 48 cards (to avoid four of a kind).
  • Choose the second remaining card from 44 cards (to avoid a full house).
  • Divide by 2! because the order in which we choose the last two cards doesn't matter (e.g., 4♥ then 9♥ is the same hand as 9♥ then 4♥).

👥 Two pairs

Two pairs of cards (each pair has the same value) and a third card of a distinct value.

• Number of hands: (13 choose 2) × (4 choose 2)^2 × (11 choose 1) × (4 choose 1) = 123,552
• Why:
  • Choose two values for the pairs: (13 choose 2).
  • Once chosen (e.g., King and 10), one is larger; choose 2 suits for the larger-valued pair: (4 choose 2).
  • Choose 2 suits for the smaller-valued pair: (4 choose 2).
  • Choose the value for the last card from the remaining 11 values (to avoid three of a kind): 11 ways.
  • Choose the suit of the last card: 4 ways.

🎴 Pair

Exactly two cards of the same value (different suits), and all other cards have different values.

• Number of hands: (13 choose 1) × (4 choose 2) × [(48 choose 1) × (44 choose 1) × (40 choose 1)] / 3! = 1,098,240
• Why:
  • Choose the value and 2 suits for the pair: (13 choose 1) × (4 choose 2).
  • Choose the first remaining card from 48 cards (to avoid three/four of a kind).
  • Choose the second from 44 cards (to avoid two pairs).
  • Choose the third from 40 cards (again to avoid two pairs).
  • Divide by 3! because the order of drawing the last three cards doesn't matter.
• Don't confuse: If the other three cards included another pair, it would be "two pairs"; if they matched the first pair's value, it would be three or four of a kind.

🃏 High card and subtraction principle

🃏 High card

A hand that does not fit any of the above categories.

• Number of hands: 1,302,540
• Combinatorial proof: Left as an exercise in the excerpt.
• Method hint (from exercises): Define set A = hands with at least two cards of the same number, B = hands where all cards are the same suit, C = all straights. Then high cards = complement of A ∪ B ∪ C. Use the Inclusion-Exclusion Principle.

➖ Why subtraction is crucial

• Many counting methods initially produce a superset that includes higher-ranking hands.
• Example: Counting flushes by choosing a suit and 5 values includes straight flushes, so we subtract 40.
• Example: Counting straights by choosing a starting value and suits for each card includes straight/royal flushes, so we subtract 40.
• Common pattern: Count broadly, then subtract the overlapping "better" hands already counted elsewhere.

🔄 Order doesn't matter: dividing by factorials

🔄 When and why to divide

• Key principle: If we choose multiple cards one at a time, but the order of selection doesn't affect the final hand, we must divide by the number of orderings (factorial).
• Three of a kind: We choose the last two cards sequentially (48, then 44), but drawing them in either order gives the same hand → divide by 2!.
• Pair: We choose the last three cards sequentially (48, 44, 40), but any ordering of these three gives the same hand → divide by 3!.
• Don't confuse: We do not divide when the order of selection corresponds to a meaningful distinction (e.g., choosing the larger vs. smaller pair in "two pairs").

🔄 Example: Three of a kind

• Without dividing: choosing 4♥ first and 9♥ second is counted separately from choosing 9♥ first and 4♥ second.
• With dividing by 2!: these two sequences are recognized as the same hand, so we divide the count by 2.

📋 Summary table of hand counts

Hand	Definition	Count	Key subtraction or division
Royal flush	A, K, Q, J, 10 of same suit	4	None (values fixed)
Straight flush	Run of 5, same suit (not royal)	36	Subtract 4 royal flushes
Four of a kind	Four same value + any card	624	None
Full house	Triple + pair	3,744	None
Flush	Same suit, not a run	5,108	Subtract 40 (straight/royal flushes)
Straight	Run of 5, ≥2 suits	10,200	Subtract 40 (straight/royal flushes)
Three of a kind	Triple + 2 distinct others	54,912	Divide by 2! (order of last 2 cards)
Two pairs	Two pairs + 1 other	123,552	None (larger/smaller pair distinguished)
Pair	One pair + 3 distinct others	1,098,240	Divide by 3! (order of last 3 cards)
High card	None of the above	1,302,540	Use Inclusion-Exclusion (exercise)

🧭 Overview

🧠 One-sentence thesis

Pascal's triangle provides a fast recursive method to compute binomial coefficients, where each entry equals the sum of the two entries above it, and these coefficients appear as the multipliers in binomial expansions.

📌 Key points (3–5)

• What Pascal's triangle is: an infinite triangular chart where the entry in the n-th row and k-th column is the binomial coefficient "n choose k."
• How to compute entries quickly: edges are always 1; interior entries equal the sum of the two entries directly above (left and right).
• Physical interpretation: each entry counts the number of distinct paths a ball can take to reach that position when falling through the grid (moving down-left or down-right at each step).
• Common confusion: the triangle can be used both as a lookup table for binomial coefficients and as a path-counting tool—these are the same numbers but viewed differently.
• Why it matters: Pascal's triangle gives the coefficients in the Binomial Theorem, connecting combinatorics to algebra.

🔢 Structure and notation

🔢 How the triangle is organized

• The triangle is arranged in rows numbered n = 0, 1, 2, 3, ...
• Within each row, entries are indexed by k = 0, 1, 2, ..., n (diagonal columns).
• The entry in row n and column k is the binomial coefficient "n choose k," written as (n k).
• Example: Row n = 4 contains the entries (4 0), (4 1), (4 2), (4 3), (4 4).

📐 Visual layout

The first few rows look like this:

• n = 0: 1
• n = 1: 1 1
• n = 2: 1 2 1
• n = 3: 1 3 3 1
• n = 4: 1 4 6 4 1
• n = 5: 1 5 10 10 5 1
• n = 6: 1 6 15 20 15 6 1

The triangle is symmetric: the left and right edges are both 1, and entries mirror across the center.

🧮 Pascal's Recurrence

🧮 The recursive rule (Theorem 4.1.1)

Pascal's Recurrence: For any natural number n, (a) the edge entries satisfy (n 0) = 1 and (n n) = 1, and (b) if 0 ≤ k < n, then (n k) + (n k+1) = (n+1 k+1).

• Part (a) says all entries on the left and right edges of the triangle are 1.
• Part (b) says any interior entry is the sum of the two entries immediately above it (one to the left, one to the right).
• Example: To find the middle entry in row n = 2, compute 2 = 1 + 1. To find the third entry (k = 2) in row n = 6, compute 15 = 5 + 10.

🔍 Why this works (algebraic proof)

The excerpt proves part (b) by manipulating factorials:

• Start with the formula for binomial coefficients: (n k) = n! / (k!(n - k)!).
• Add (n k) + (n k+1) by finding a common denominator.
• Multiply the first term by (k+1)/(k+1) and the second by (n - k)/(n - k).
• Combine to get n!(k + 1 + n - k) / ((k+1)!(n - k)!) = n!(n+1) / ((k+1)!(n - k)!).
• Simplify to (n+1)! / ((k+1)!(n+1 - (k+1))!) = (n+1 k+1).

Don't confuse: This is an algebraic proof; the excerpt also mentions a "more elegant combinatorial proof" in a later section (Example 5.3.3).

🎯 Path-counting interpretation

🎯 The falling ball model (Lemma 4.1.2)

Lemma 4.1.2: The entry (n k) counts the number of possible ways a ball could have traveled to that location in the grid.

• Imagine a ball starting at the top of the triangle.
• At each step, the ball falls down one row, moving either left (L) or right (R).
• To reach row n, the ball makes n moves.
• To reach the entry (n k), the ball must move right exactly k times and left n - k times.
• The number of such routes is (n k), because you choose which k of the n steps are "right."

🛤️ Examples of routes

• Example 4.1.3: To reach (4 0), there is only one route: LLLL. To reach (4 1), there are four routes: LLLR, LLRL, LRLL, RLLL.
• Example 4.1.4 (grid walking): Starting at a corner, walk two blocks north and three blocks west to reach a destination. After rotating Pascal's triangle, the number of such paths is 10. The paths can be written as sequences like NNWWW, NWNWW, etc.

🧩 Why recurrence makes sense for paths

• Part (a): There is only one way to reach the outer edges—always move left or always move right.
• Part (b): To reach an interior spot, the ball must have come from one of two positions in the row above (either from the left or from the right). The total number of paths is the sum of paths to those two positions.

🔗 Connection to the Binomial Theorem

🔗 Binomial coefficients in algebra (Theorem 4.2.2)

Binomial Theorem: The expansion of (x + y) raised to the power n is (x + y)^n = (n 0)x^n + (n 1)x^(n-1)y^1 + ... + (n k)x^(n-k)y^k + ... + (n n)y^n.

• The coefficients in the expansion are exactly the entries in row n of Pascal's triangle.
• Example expansions:
  • (x + y)^0 = 1
  • (x + y)^1 = x + y
  • (x + y)^2 = x^2 + 2xy + y^2
  • (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
  • (x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4

🧪 Why the coefficients match (proof sketch)

• Expanding (x + y)^n means multiplying (x + y) by itself n times.
• At each of the n factors, you choose either x or y.
• To get the term x^(n-k) y^k, you must choose y exactly k times and x the remaining n - k times.
• There are (n k) ways to choose which k factors contribute y.
• This is the "giant foil method" generalization mentioned in Remark 4.2.1.

Don't confuse: The binomial coefficients are the same numbers whether you compute them via factorials, Pascal's recurrence, path counting, or algebraic expansion—they are all equivalent views of the same combinatorial object.

🧪 Practical uses and exercises

🧪 Computing rows

• To find row n = 7 or n = 8, start from row n = 6 and apply Pascal's recurrence repeatedly.
• Each new row has one more entry than the previous row.

🧪 Grid path problems

• General grid paths: The number of paths of length m + n from the lower left to the upper right corner of an m × n grid (moving only along edges, not backtracking) is a binomial coefficient.
• Example: Walking four blocks north and two blocks west (without going south or east) corresponds to choosing which 2 of the 6 total steps are "west."

🧪 Probability and symmetry

• If a ball falls with equal probability of going left or right at each step, the probability of landing at a specific entry depends on the number of paths to that entry divided by the total number of paths (which is 2^n for row n).
• Example (Exercise 9): Compute the probability the ball lands at the 3rd spot in the 5th row.
• Symmetry observation (Exercise 10): In row 10, the ball has a 50% probability of landing where k is even, because the triangle is symmetric and paths are equally likely.

🧪 Divisibility property (Exercise 6)

• If p is a prime number and 1 ≤ k ≤ p - 1, then (p k) is divisible by p (i.e., (p k) ≡ 0 mod p).
• This follows from the factorial formula and properties of primes.

🧭 Overview

🧠 One-sentence thesis

The Binomial Theorem states that the coefficients in the expansion of (x + y) to the power n are precisely the binomial coefficients from Pascal's triangle, which can be explained by counting how many ways to choose variables from each factor.

📌 Key points (3–5)

• The pattern: When you expand (x + y) to the power n, the coefficients are the numbers in row n of Pascal's triangle—this is why they are called "binomial coefficients."
• The formula: The expansion is the sum of terms (n choose k) times x to the (n minus k) times y to the k, for k from 0 to n.
• Why it works: To get a term like x to the (n minus k) times y to the k, you must choose y from exactly k of the n factors (x + y), and there are (n choose k) ways to do that.
• Common confusion: The "foil" method for (x + y) squared is a special case; the Binomial Theorem is the "giant foil method" for any power.
• Practical use: By reading a row from Pascal's triangle, you can immediately write down the expansion of any binomial power without multiplying everything out.

🔢 The pattern in powers of binomials

🔢 Observing the coefficients

The excerpt shows the first few powers of (x + y):

• (x + y) to the 0 equals 1
• (x + y) to the 1 equals x + y
• (x + y) to the 2 equals x squared + 2xy + y squared
• (x + y) to the 3 equals x cubed + 3 x squared y + 3 x y squared + y cubed
• (x + y) to the 4 equals x to the 4 + 4 x cubed y + 6 x squared y squared + 4 x y cubed + y to the 4

When you ignore the x's and y's and look only at the coefficients (1; 1, 1; 1, 2, 1; 1, 3, 3, 1; 1, 4, 6, 4, 1), they are exactly the rows of Pascal's triangle.

📛 Why "binomial coefficients"

The numbers (n choose k) are called binomial coefficients because they appear as the coefficients in the expansion of a binomial (x + y) to the power n.

This naming convention directly reflects the pattern observed above.

🧮 The Binomial Theorem statement and proof

🧮 The theorem

Binomial Theorem: Let x and y be variables. Then the expansion of (x + y) to the power n is: (x + y) to the n equals (n choose 0) x to the n plus (n choose 1) x to the (n minus 1) y to the 1 plus ... plus (n choose k) x to the (n minus k) y to the k plus ... plus (n choose n minus 1) x to the 1 y to the (n minus 1) plus (n choose n) y to the n.

• This theorem was first discovered by the Persian mathematician and engineer Al-Karaji (approximately 953–1029), making it one of the oldest named theorems in mathematics.
• Each term has the form (n choose k) times x to the (n minus k) times y to the k.
• The powers of x decrease from n to 0, while the powers of y increase from 0 to n; the sum of the exponents in each term is always n.

🔍 Why the theorem works (the "giant foil method")

The proof explains the counting argument:

1. Start with the product: (x + y) to the n means multiplying (x + y) by itself n times.
2. Expand by choosing: To expand, you take one entry (either x or y) from each of the n factors, multiply them together, and add all possible products.
3. Count the ways to get a specific term: To produce the monomial x to the (n minus k) times y to the k, you must choose y from exactly k of the n factors and x from the remaining (n minus k) factors.
4. Apply the binomial coefficient: There are (n choose k) ways to choose which k factors contribute y (and which n minus k contribute x), so the coefficient of x to the (n minus k) times y to the k is (n choose k).

Example: In (x + y) to the 10, the coefficient in front of x to the 8 times y squared is (10 choose 2) equals 45, because you need to choose y from 2 of the 10 factors (and x from the remaining 8).

🧩 Connection to "foil"

The excerpt notes that some students know the "foil" (first, outer, inner, last) method for (x + y) squared:

• (x + y) squared equals x times x plus y times x plus x times y plus y times y.

The Binomial Theorem generalizes this into a "giant foil method" for any power n, systematically counting all the ways to pick x or y from each factor.

🛠️ Using the theorem with Pascal's triangle

🛠️ Quick expansion by reading rows

By combining the Binomial Theorem with Pascal's triangle, you can expand powers without multiplying:

Example 1: Row n equals 6 of Pascal's triangle is 1, 6, 15, 20, 15, 6, 1, so:

• (x + y) to the 6 equals x to the 6 plus 6 x to the 5 y plus 15 x to the 4 y squared plus 20 x cubed y cubed plus 15 x squared y to the 4 plus 6 x y to the 5 plus y to the 6.

Example 2: Row n equals 3 is 1, 3, 3, 1, so:

• (a squared + 5b) to the 3 equals (a squared) cubed plus 3 times (a squared) squared times (5b) plus 3 times (a squared) times (5b) squared plus (5b) cubed
• Simplifying: a to the 6 plus 15 a to the 4 b plus 75 a squared b squared plus 125 b cubed.

🎯 Finding a single coefficient

You don't need to expand the entire expression to find one coefficient—just use the appropriate binomial coefficient.

Example: What is the coefficient of x to the 5 times y cubed in (x + y) to the 8?

• The term x to the 5 times y cubed corresponds to k equals 3 (since y appears 3 times).
• The coefficient is (8 choose 3).

Don't confuse: the exponent on y tells you which binomial coefficient to use (k), not the exponent on x.

📐 Additional patterns in Pascal's triangle

📐 Symmetry

The excerpt mentions that (n choose k) equals (n choose n minus k), which is illustrated by the reflective symmetry about the vertical axis in Pascal's triangle.

Row	Entries	Symmetry
n = 4	1, 4, 6, 4, 1	Reads the same forwards and backwards
n = 5	1, 5, 10, 10, 5, 1	Reads the same forwards and backwards

📐 Row sums are powers of 2

The sum of all entries in row n equals 2 to the n.

Example: Row n equals 4 is 1, 4, 6, 4, 1, and 1 + 4 + 6 + 4 + 1 = 16 = 2 to the 4.

In general, the sum of the entries in row n is 2 to the n.

🧭 Overview

🧠 One-sentence thesis

Pascal's triangle exhibits three fundamental patterns—symmetry, row sums equaling powers of 2, and alternating row sums equaling zero—all of which can be proven using the Binomial Theorem and interpreted through a falling-ball probability model.

📌 Key points (3–5)

• Symmetry pattern: The triangle is symmetric about its vertical axis, reflecting the identity that binomial coefficient (n choose k) equals (n choose n minus k).
• Row sums pattern: The sum of all entries in row n equals 2 to the power n, proven by substituting x equals 1 and y equals 1 into the Binomial Theorem.
• Alternating sums pattern: For positive n, the alternating sum (adding and subtracting consecutive entries) in row n equals zero, proven by substituting x equals 1 and y equals negative 1 into the Binomial Theorem.
• Common confusion: The falling-ball probability interpretation connects combinatorial counting to probability—each binomial coefficient (n choose k) divided by 2 to the power n gives the probability of reaching that location, not just a count.
• Why it matters: These patterns provide multiple ways to understand binomial coefficients—geometric (symmetry), algebraic (Binomial Theorem), and probabilistic (falling ball).

🔄 The symmetry pattern

🪞 What symmetry means

Symmetry in Pascal's triangle: The triangle is reflectively symmetric about its vertical axis.

• This reflects the identity: (n choose k) equals (n choose n minus k).
• The excerpt references Corollary 3.4.5 as the source of this identity.
• Example: In row n equals 4, the entry (4 choose 1) equals 4 and (4 choose 3) equals 4; both are equidistant from the center.

🔍 Visual interpretation

• The vertical axis passes through the center entries: 1, 2, 6, 20, and so on.
• Each entry at position k from the left mirrors the entry at position n minus k from the left.
• Don't confuse: Symmetry is about position within a row, not about comparing different rows.

➕ Row sums equal powers of 2

📐 The pattern statement

Row sum pattern: The sum of all entries in row n of Pascal's triangle equals 2 to the power n.

• In binomial coefficient notation: (n choose 0) plus (n choose 1) plus (n choose 2) plus ... plus (n choose n) equals 2 to the power n.
• Example: For n equals 4, the sum is 1 plus 4 plus 6 plus 4 plus 1 equals 16 equals 2 to the power 4.

🧮 Proof using the Binomial Theorem

• The Binomial Theorem states: (x plus y) to the power n equals the sum over k from 0 to n of (n choose k) times x to the power (n minus k) times y to the power k.
• Set x equals 1 and y equals 1: (1 plus 1) to the power n equals the sum of (n choose k) times 1 to the power (n minus k) times 1 to the power k.
• Simplify: 2 to the power n equals the sum of all (n choose k) from k equals 0 to n.
• This proves Proposition 4.3.1.

🎲 Falling-ball probability interpretation

• Imagine a ball falling through Pascal's triangle, with equal probability (1/2) of going left or right at each step.
• After n steps, the ball has taken one of 2 to the power n possible routes, each with probability 1 divided by 2 to the power n.
• The number of routes passing through location (n choose k) is exactly (n choose k).
• Therefore, the probability of passing through (n choose k) is (n choose k) divided by 2 to the power n (Lemma 4.3.2).

🔗 Connecting probability to row sums

• The ball must land somewhere in row n after n steps, so the sum of all probabilities equals 1.
• Multiply each probability by 2 to the power n: the sum of all (n choose k) equals 2 to the power n.
• This reproduces the row sum pattern (Remark 4.3.3).
• Example: For n equals 4, there is 1 path to (4 choose 0) with probability 1/16; there are 4 paths to (4 choose 1) with probability 4/16 equals 1/4 (Example 4.3.4).

➖ Alternating sums equal zero

📐 The alternating sum pattern

Alternating sum pattern: For positive n, the alternating sum (n choose 0) minus (n choose 1) plus (n choose 2) minus (n choose 3) plus ... equals 0.

• The sign alternates between plus and minus.
• The last term is positive if n is even, negative if n is odd.
• Example: For n equals 4, the alternating sum is 1 minus 4 plus 6 minus 4 plus 1 equals 0.

🧮 Proof using the Binomial Theorem

• Set x equals 1 and y equals negative 1 in the Binomial Theorem: (1 plus (negative 1)) to the power n equals the sum of (n choose k) times 1 to the power (n minus k) times (negative 1) to the power k.
• Simplify: 0 to the power n equals (n choose 0) minus (n choose 1) plus (n choose 2) minus (n choose 3) plus ... (with alternating signs).
• For positive n, 0 to the power n equals 0, proving Proposition 4.3.5.
• The excerpt notes that another proof will be given in Proposition 5.4.8.

📊 Summary of the three patterns

Pattern	Formula	Proof method	Interpretation
Symmetry	(n choose k) equals (n choose n minus k)	Corollary 3.4.5	Reflective symmetry about vertical axis
Row sums	Sum of (n choose k) for k from 0 to n equals 2 to the power n	Binomial Theorem with x equals 1, y equals 1	Total probability equals 1; ball must land somewhere
Alternating sums	Alternating sum of (n choose k) equals 0 (for positive n)	Binomial Theorem with x equals 1, y equals negative 1	Cancellation of positive and negative contributions

🎯 How to distinguish the patterns

• Row sums vs alternating sums: Row sums add all entries with the same sign; alternating sums change the sign at each step.
• Symmetry vs sums: Symmetry compares two entries within the same row; sums combine all entries in a row into a single number.
• Binomial Theorem substitutions: Different substitutions (x equals 1, y equals 1 versus x equals 1, y equals negative 1) yield different patterns.

🧭 Overview

🧠 One-sentence thesis

Beyond basic patterns, Pascal's triangle contains deeper identities such as the sum-of-squares formula and the hockey stick identity, which reveal hidden relationships between binomial coefficients.

📌 Key points (3–5)

• Sum of squares pattern: the sum of the squares of all entries in row n equals the middle entry of row 2n.
• Hockey stick identity: a diagonal sum of binomial coefficients along a "hockey stick" shape equals a single binomial coefficient at the end.
• Two versions of hockey stick: version 1 starts from the top diagonal, version 2 starts from a fixed column; they are related by symmetry.
• Common confusion: the hockey stick identity involves entries from multiple rows, not just one row, and the sum traces a diagonal path.
• Why it matters: these identities connect different parts of Pascal's triangle and will be proved later using combinatorial arguments.

🔢 Sum of squares identity

🔢 What the pattern says

Proposition 4.4.1 (Sum of squares): If n is a natural number, then (n choose 0)² + (n choose 1)² + ... + (n choose n)² = (2n choose n).

• This identity states that if you square every entry in row n and add them up, the result is the middle entry of row 2n.
• Example: for n = 2, we have 1² + 2² + 1² = 6, which is the middle entry of row 4.

📊 Numerical examples

The excerpt lists:

Row n	Sum of squares	Result	Where it appears
0	1²	1	Middle of row 0
1	1² + 1²	2	Middle of row 2
2	1² + 2² + 1²	6	Middle of row 4
3	1² + 3² + 3² + 1²	20	Middle of row 6
4	1² + 4² + 6² + 4² + 1²	70	Middle of row 8

• The sequence 1, 2, 6, 20, 70, ... appears as the middle entries in even-numbered rows.
• Don't confuse: this is not about adding entries in one row; it's about squaring each entry first, then summing.

🔮 Proof deferred

• The excerpt states that Proposition 4.4.1 will be proved later in Proposition 5.3.4.
• The proof is not given here; the section focuses on recognizing the pattern.

🏒 Hockey stick identity

🏒 What the pattern says

Proposition 4.4.2 (Hockey stick identity): Suppose n is a positive integer and m is a non-negative integer.

• Version 1: (n choose 0) + (n+1 choose 1) + (n+2 choose 2) + ... + (n+m choose m) = (n+m+1 choose m).
• Version 2: (n choose n) + (n+1 choose n) + (n+2 choose n) + ... + (n+m choose n) = (n+m+1 choose n+1).

• The name comes from the visual shape: the binomial coefficients involved trace a diagonal path resembling a hockey stick in Pascal's triangle.
• Version 2 is obtained from version 1 by applying the symmetry property to each binomial coefficient.

🎯 How to read the identity

• Version 1: start at (n choose 0), move diagonally down-right, adding entries until you reach (n+m choose m); the sum equals the entry just below and to the right of the last term.
• Version 2: start at (n choose n), move straight down in the same column, adding entries; the sum equals an entry one row below and one column to the right.
• Don't confuse: the sum involves entries from different rows, not a single row.

🖼️ Visual example

The excerpt illustrates version 1 with n = 2 and m = 4:

• Start at row 2, position 0: entry 1 (which is (2 choose 0)).
• Move diagonally: row 3, position 1: entry 3 (which is (3 choose 1)).
• Continue: row 4, position 2: entry 6; row 5, position 3: entry 10; row 6, position 4: entry 15.
• Sum: 1 + 3 + 6 + 10 + 15 = 35, which equals (7 choose 4).

The excerpt shows this path circled in Pascal's triangle:

n = 0:   1
n = 1:   1   1
n = 2:   1   2   1       ← start here (1)
n = 3:   1   3   3   1   ← next (3)
n = 4:   1   4   6   4   1   ← next (6)
n = 5:   1   5   10  10  5   1   ← next (10)
n = 6:   1   6   15  20  15  6   1   ← next (15)
n = 7:   1   7   21  35  35  21  7   1   ← result (35)

🔮 Proof deferred

• The excerpt states that Proposition 4.4.2 will be proved later in Proposition 5.2.10.
• One exercise hints at a combinatorial proof: separate the 35 paths ending at (7 choose 4) into sets of sizes 15, 10, 6, 3, 1 using Lemma 4.1.2 (paths interpretation).

🧪 Exercises and applications

🧪 Drawing and verifying

• Exercise 1 asks to draw version 2 of the hockey stick identity for n = 2, m = 4 on Pascal's triangle.
• Exercise 2 asks to explain version 1 using path-counting (Lemma 4.1.2): the 35 paths to (7 choose 4) can be grouped by where they pass through earlier entries.

📈 Chi-square distribution application

• Exercise 3 introduces the chi-square distribution: Q_n = x₀² + ... + xₙ² for the entries in row n.
• Part (a): use Proposition 4.4.1 to compute Q_n for n = 1, 2, 3, 4, 5 (these are the middle entries of rows 2, 4, 6, 8, 10).
• Part (b) and (c): ask what happens to Q_n as n grows, and what the limit is as n approaches infinity.
• Don't confuse: this is an application of the sum-of-squares identity to probability theory, not a new identity.

🧭 Overview

🧠 One-sentence thesis

Multinomial coefficients count the number of distinct anagrams of a word by dividing the total permutations by the factorial product of each repeated letter's count.

📌 Key points (3–5)

• What anagrams are: rearrangements of letters in a word where some letters may repeat (different from permutations where all elements are distinct).
• Why simple permutation counting fails: when letters repeat, the standard n! formula overcounts because swapping identical letters produces the same anagram.
• The multinomial coefficient formula: for a word with n letters of m different types (with k₁, k₂, ..., kₘ occurrences), the number of anagrams is n! divided by (k₁! · k₂! · ... · kₘ!).
• Common confusion: permutations vs anagrams—permutations treat every position as distinct; anagrams recognize that identical letters are interchangeable, so fewer distinct arrangements exist.
• Special case: when m = 2 (only two letter types), the multinomial coefficient reduces to the familiar binomial coefficient.

🔤 What makes anagrams different from permutations

🔤 Anagrams defined

An anagram is a rearrangement of the letters in a word.

• Unlike permutations, anagrams account for repeated letters.
• When letters repeat, many permutations represent the same anagram.
• Example: the word ZOO has two O's; if we label them O₁ and O₂, we get 3! = 6 permutations (ZO₁O₂, ZO₂O₁, O₁ZO₂, O₂ZO₁, O₁O₂Z, O₂O₁Z), but only 3 distinct anagrams (ZOO, OZO, OOZ) because swapping the two O's doesn't create a new anagram.

🔄 Why repetition causes overcounting

• Standard permutation counting assumes all elements are distinct.
• When k identical letters exist, each anagram appears k! times in the full permutation list.
• Example: SASSY has 5 letters, so 5! = 120 permutations, but three S's can be rearranged in 3! = 6 ways without changing the anagram, so the actual number of anagrams is 5! / 3! = 20.

🧮 The multinomial coefficient

🧮 Definition and notation

Let n be a natural number. Let k₁, ..., kₘ be m positive natural numbers such that k₁ + ... + kₘ = n. Let W be a word with n letters of m different types, in which the letter of type i appears kᵢ times, for 1 ≤ i ≤ m. The multinomial coefficient (n choose k₁, ..., kₘ) is defined to be the number of anagrams of W.

• The notation is (n choose k₁, ..., kₘ).
• It generalizes the binomial coefficient to more than two types of elements.
• The k values must sum to n (total letter count).

📐 The formula

The multinomial coefficient formula is:

(n choose k₁, ..., kₘ) = n! / (k₁! · k₂! · ... · kₘ!)

• Start with n! (all permutations if letters were distinct).
• Divide by each kᵢ! to correct for the overcounting caused by each letter type's repetitions.
• Example: FORTCOLLINS has 11 letters with two L's and two O's (and 7 other distinct letters), so the count is 11! / (2! · 2! · 1! · 1! · 1! · 1! · 1! · 1! · 1!) = 11! / 4.

🔢 Worked example: MISSISSIPPI

• Total letters: n = 11
• Letter types: m = 4
• Counts: k₁ = 4 (I's), k₂ = 1 (M), k₃ = 2 (P's), k₄ = 4 (S's)
• Number of anagrams: 11! / (4! · 1! · 2! · 4!)

Don't confuse: the denominator includes a factorial for every letter type, even if it appears only once (like the M, which contributes 1! = 1).

🔗 Connection to binomial coefficients

🔗 Special case when m = 2

• When there are only two letter types, the multinomial coefficient becomes the binomial coefficient.
• If k₁ + k₂ = n, then (n choose k₁, k₂) = n! / (k₁! · k₂!) = n! / (k₁! · (n − k₁)!) = (n choose k₁).
• Example: a binary string of length n with k zeros and (n − k) ones has (n choose k) anagrams, because the anagram is determined exactly by where the k zeros are placed.

🔗 Why the formulas match

• The binomial coefficient formula n! / (k₁! · (n − k₁)!) from earlier theorems is identical to the multinomial formula when m = 2.
• This shows that multinomial coefficients are a true generalization of binomial coefficients.

🌐 Applications beyond word anagrams

🌐 Paths in higher dimensions

• The excerpt mentions counting paths from (0, 0, 0) to (3, 4, 5) in three-dimensional space using only unit steps in directions (1, 0, 0), (0, 1, 0), and (0, 0, 1).
• This is equivalent to counting anagrams of a "word" with 3 steps of type 1, 4 steps of type 2, and 5 steps of type 3.
• Total steps: n = 3 + 4 + 5 = 12; answer: 12! / (3! · 4! · 5!).

🌐 Ordered set partitions preview

• The excerpt briefly introduces ordered set partitions as a generalization.
• Example: choosing 2 fruits on Monday, 3 on Tuesday, and 2 on Wednesday from 7 fruits gives (7 choose 2) · (5 choose 3) · (2 choose 2) = 7! / (2! · 3! · 2!) = 210 ways.
• This is the same structure as the multinomial coefficient: dividing a set into ordered groups of specified sizes.

🧭 Overview

🧠 One-sentence thesis

Ordered set partitions generalize the binomial coefficient by counting ways to separate a set into more than two labeled parts, and the multinomial coefficient formula directly extends the binomial theorem to powers of sums with more than two terms.

📌 Key points (3–5)

• What ordered set partitions count: ways to separate a set into multiple non-overlapping labeled blocks (not just two parts like binomial coefficients).
• The multinomial coefficient formula: n! divided by the product of all block sizes' factorials, counting ordered set partitions.
• Connection to anagrams: anagrams of a word with repeated letters correspond exactly to ordered set partitions of position indices.
• Common confusion: ordered set partitions vs. set partitions—ordered partitions assign labels/roles to blocks (e.g., Monday/Tuesday/Wednesday), while set partitions treat all blocks as interchangeable.
• Why it matters: the multinomial theorem generalizes the binomial theorem, giving coefficients for expanding powers of sums with more than two variables.

🧩 Core concept: ordered set partitions

🧩 What an ordered set partition is

Ordered set partition: a collection of non-empty subsets B₁, ..., Bₘ of a set S such that any two blocks have trivial intersection and their union is S; the blocks are ordered (labeled).

• This generalizes the binomial coefficient idea: instead of splitting S into "chosen" and "not chosen" (two parts), we split S into m labeled parts.
• The blocks are ordered, meaning B₁ is distinct from B₂ even if they have the same size.
• Example: Splitting 7 fruits into "Monday's 2 fruits," "Tuesday's 3 fruits," and "Wednesday's 2 fruits" creates an ordered set partition with three blocks.

🔢 The multinomial coefficient

Multinomial coefficient: (n choose k₁, k₂, ..., kₘ) = n! / (k₁! · k₂! · ... · kₘ!), where k₁ + k₂ + ... + kₘ = n.

• This counts the number of ordered set partitions of an n-element set into blocks of sizes k₁, k₂, ..., kₘ.
• The proof works by successive counting: choose k₁ elements for the first block (n choose k₁ ways), then k₂ from the remaining (n−k₁ choose k₂ ways), and so on; all intermediate factorials cancel, leaving only n! in the numerator and the block-size factorials in the denominator.
• Example: 7 fruits into blocks of sizes 2, 3, 2 gives (7 choose 2,3,2) = 7!/(2!·3!·2!) = 210.

🎁 Example: distributing presents

• 11 different presents to 4 people in order: 4 to person a, 1 to person b, 2 to person c, 4 to person d.
• This is an ordered set partition with n=11 and block sizes 4, 1, 2, 4.
• Number of ways = (11 choose 4,1,2,4) = 11!/(4!·1!·2!·4!) = 34,650.

🔤 Connection to anagrams

🔤 How anagrams correspond to ordered set partitions

• An anagram of a word with repeated letters can be encoded as an ordered set partition of position indices {1, 2, ..., n}.
• Given an anagram, for each distinct letter i (in alphabetical order), collect the positions where that letter appears into block Bᵢ.
• Example: the anagram IMSSSPIIPSI of MISSISSIPPI has I's in positions {1,7,8,11}, M in {2}, P's in {5,9}, S's in {3,4,6,10}; this gives the ordered set partition with those four blocks.
• Conversely, given an ordered set partition with blocks of sizes matching the letter counts, place the i-th letter in the positions numbered by block Bᵢ to reconstruct an anagram.

🧮 Proof of the anagram counting formula

• By the correspondence above, the number of anagrams of a word W with k₁ copies of letter a₁, k₂ copies of a₂, ..., kₘ copies of aₘ equals the number of ordered set partitions of {1, ..., n} into blocks of sizes k₁, ..., kₘ.
• By the multinomial coefficient theorem, this count is n! / (k₁! · k₂! · ... · kₘ!).
• This proves Theorem 4.5.7 (the anagram counting formula).

🆚 Ordered vs. unordered set partitions

🆚 The key distinction

Feature	Ordered set partition	Set partition
Blocks labeled?	Yes—blocks have distinct roles (e.g., Monday/Tuesday/Wednesday or person a/b/c/d)	No—blocks are interchangeable
Counting formula	(n choose k₁,...,kₘ) = n!/(k₁!·...·kₘ!)	Divide the ordered count by the number of ways to permute identical-sized blocks

• Don't confuse: if blocks have assigned roles or labels, use ordered set partitions; if blocks are just "groups" with no distinction, use set partitions.

🎓 Example: students in groups

• Problem 1 (ordered): 12 students separated into 4 groups of 3, assigned to present on Monday, Tuesday, Wednesday, Thursday.
  • Each assignment is an ordered set partition into blocks B₁, B₂, B₃, B₄ of size 3.
  • Answer: (12 choose 3,3,3,3) = 12!/(3!·3!·3!·3!).
• Problem 2 (unordered): 12 students separated into 4 groups of 3, with no day assignment.
  • This is a set partition; the 4 blocks are interchangeable.
  • There are 4! ways to order the 4 blocks, so divide the ordered count by 4!.
  • Answer: 12!/(3!·3!·3!·3! · 4!).

🧪 The multinomial theorem

🧪 Statement of the theorem

Multinomial Theorem: The expansion of (x₁ + x₂ + ... + xₘ)ⁿ is the sum over all k₁ + ... + kₘ = n of (n choose k₁,...,kₘ) · x₁^k₁ · ... · xₘ^kₘ.

• This directly generalizes the binomial theorem (which handles m=2) to sums of more than two variables.
• The coefficient of any monomial x₁^k₁ · ... · xₘ^kₘ in the expansion is the multinomial coefficient (n choose k₁,...,kₘ).
• Example: (x + y + z)³ = x³ + y³ + z³ + 3x²y + 3xy² + 3x²z + 3xz² + 3y²z + 3yz² + 6xyz; the coefficient 3 of x²y¹z⁰ equals (3 choose 2,1,0) = 3.

🧪 Why the multinomial coefficient appears

• To form the monomial x₁^k₁ · ... · xₘ^kₘ in the product (x₁ + ... + xₘ)ⁿ, label the n factors as F₁, ..., Fₙ.
• Choose k₁ factors from which to pick x₁, then k₂ factors from which to pick x₂, and so on.
• This creates an ordered set partition of the set of factors {F₁, ..., Fₙ} into blocks of sizes k₁, ..., kₘ.
• By Theorem 4.6.3, the count is (n choose k₁,...,kₘ), which is the coefficient of that monomial.

🎯 Practice problems

🎯 Distributing distinct objects

• Pillows to students: 20 different pillows to 12 students (a through l), with 2 pillows each to students a–h and 1 pillow each to students i–l.

  • This is an ordered set partition with n=20 and block sizes 2,2,2,2,2,2,2,2,1,1,1,1.
  • Answer: (20 choose 2,2,2,2,2,2,2,2,1,1,1,1) = 20!/(2!⁸ · 1!⁴).

• Months to nurses: 12 months assigned to 3 nurses (a, b, c), 4 months each.

  • Ordered set partition with n=12 and block sizes 4,4,4.
  • Answer: (12 choose 4,4,4) = 12!/(4!·4!·4!).

🎯 Multi-stage problems

• Conference room assignments: 9 friends in 3 hotel rooms (sizes 4, 3, 2) for 5 nights; room assignments can change each night.

  • Each night is an independent ordered set partition of 9 people into blocks of sizes 4, 3, 2.
  • For one night: (9 choose 4,3,2) = 9!/(4!·3!·2!).
  • For 5 nights: raise this to the 5th power (since each night is independent).
  • Answer: [9!/(4!·3!·2!)]⁵.

• Stickers on windows: 8 different stickers on 4 windows, 2 stickers per window.

  • Ordered set partition with n=8 and block sizes 2,2,2,2.
  • Answer: (8 choose 2,2,2,2) = 8!/(2!·2!·2!·2!).

🧭 Overview

🧠 One-sentence thesis

This investigation uses the chessboard as a structured setting to explore counting problems involving placement constraints, attacking relationships, and material distinctions among chess pieces.

📌 Key points (3–5)

• The setting: an 8×8 chessboard (or n×n generalization) with pieces that move according to chess rules; the board cannot be rotated or flipped.
• Core constraint types: no two pieces on the same square, no attacking between pieces of different materials, and maximizing the number of pieces under constraints.
• Rook problems: focus on row/column restrictions (rooks attack along rows and columns).
• Queen problems: more complex because queens attack along rows, columns, and diagonals.
• Common confusion: "attacking" only applies between pieces of different materials; pieces of the same material can be placed to avoid attacking each other, but the constraint changes when materials differ.

♟️ The chessboard framework

♟️ Board structure

A chessboard is an 8×8 board tiled in 1×1 squares, colored black and white so that no black square shares an edge with a white square.

• The board is fixed to the table—it cannot be rotated or turned over.
• This means positions are absolute; rotating the board would create a different configuration.

🎯 Piece movement and attacking

• Rook: moves to any square in the same row or column.
• Queen: moves to any square in the same row, column, or diagonal.
• Attacking definition: two pieces of different materials attack each other if one can move to the other's square in one move.
• Example: a gold rook and a silver rook attack each other if they share a row or column; two gold rooks do not "attack" in the sense used here.

🧱 Material distinctions

• Problems involve pieces made from different materials (gold, silver, titanium, diamond).
• An unlimited supply of rooks (or queens) of each material is available.
• The key constraint is often preventing attacks between different materials, not within the same material.

🏰 Rook placement problems

🏰 Basic rook constraints

Problem 1: Place two gold rooks and two silver rooks.

• (a) No square has more than one rook.
• (b) Additionally, no gold rook attacks a silver rook (i.e., no gold and silver rook share a row or column).

📏 Maximum non-attacking rooks (same material)

Problem 2: What is the largest number of rooks of the same material that can be placed so no two share a row or column?

• Since a rook attacks along its row and column, each rook must occupy a unique row and a unique column.
• On an 8×8 board, the maximum is 8 rooks (one per row, one per column).
• The problem also asks: how many ways can this be done?

🌈 Multi-material rook arrangements

Problem 3: Place 16 rooks (four of each of four materials) so that no pair of rooks with different materials attacks each other.

• This requires partitioning the board so that rooks of different materials never share rows or columns.
• The problem asks for a demonstration that this is possible and a count of the number of ways.

⚖️ The peaceful rooks problem

Problem 4: Find the largest number m such that m gold rooks and m silver rooks can be placed with no gold rook attacking a silver rook.

• This is a maximization problem under the constraint that the two groups of rooks do not interfere.
• The problem also asks for the number of ways to achieve this maximum.

👑 Queen placement problems

👑 Queen mobility

Problem 5: Analyze a single queen's movement options.

• (a) Minimum mobility: find a position where the queen can move to the fewest squares. (Likely a corner or edge position.)
• (b) Maximum mobility: find a position where the queen can move to the most squares. (Likely a central position.)

🔢 The n-queens puzzle

Problem 6: Place n gold queens on an n×n board so no two attack each other.

• (a) Show this is impossible for n=2 or n=3.
• (b)–(c) Find solutions for n=4 and n=5.
• (d) The "8 queens puzzle" is a famous instance; the problem asks students to look it up and draw a solution.
• Don't confuse: this is about queens of the same material not attacking each other, which is different from the multi-material constraint.

🕊️ The peaceable queens problem

Problem 7: On an n×n board, find the maximum m such that m gold queens and m silver queens can be placed with no attacks between different materials.

• Example: when n=3, m=1 (one gold queen in a corner, one silver queen in the middle of an opposite side).
• (a) For n=4, show m≥2 by constructing a valid placement.
• (b) For n=5, show m≥4.
• (c) Look up the peaceable queens problem and find m for n=8.
• This generalizes the peaceful rooks problem to the more complex queen movement pattern.

🔗 Connection to broader combinatorics

🔗 Why these problems matter

• The excerpt transitions to Chapter 5, which covers proof techniques in combinatorics.
• Chessboard problems illustrate counting under constraints, a central theme in combinatorics.
• They require systematic reasoning about placement rules, material distinctions, and optimization—skills that proofs formalize.

📚 Proofs as explanation

Proofs are valuable because they explain the reasons why statements are true.

• The excerpt emphasizes that proofs are not just verification but also understanding.
• Chessboard problems naturally lead to questions like "Why is this the maximum?" or "How many ways exist?"—questions that require proof.
• Example: proving that 8 is the maximum number of non-attacking rooks (same material) on an 8×8 board involves showing both that 8 is achievable and that 9 is impossible.

🧭 Overview

🧠 One-sentence thesis

Proofs are essential in mathematics because numerical patterns and computational checks can be misleading, and only rigorous reasoning can establish that a statement holds without exception.

📌 Key points (3–5)

• Why data misleads: patterns observed in the first few cases can break down at larger values, so checking examples is not enough.
• What proofs provide: they show a statement is true without doubt and explain why it is true, not just that it appears true.
• Historical examples: famous conjectures (Fermat numbers, Fermat's Last Theorem, Four Color Theorem) show both the power of proof and the danger of assuming patterns continue.
• Common confusion: "almost integers" and pseudo-primes look like they satisfy a property but do not—appearance is not the same as proof.
• Why proofs matter: they are the foundation of mathematics, connecting ideas across time and enabling us to build reliable knowledge.

🚫 When patterns fail

🚫 Divisors of factorials

Example 5.1.1 asks: How many positive divisors does n! have?

• The first few cases yield 1, 2, 4, 8, 16 divisors for n = 1, 2, 3, 4, 5.
• This suggests the pattern "doubles each time," so you might guess 6! has 32 divisors.
• Reality: 6! has only 30 divisors—the pattern breaks.
• Lesson: We need mathematical proofs to show a pattern continues indefinitely; we cannot rely on a finite number of checks because a counterexample may be too large to find by hand.

🎭 Almost integers and hoaxes

Example 5.1.2 presents "almost integers": numbers so close to integers that they appear to be whole numbers.

• e^(π√43) ≈ 884736743.9997
• e^(π√67) ≈ 147197952743.999998
• e^(π√163) ≈ 262537412640768743.9999999999992 (Ramanujan's constant)
• In April 1975, Gardner claimed Ramanujan conjectured e^(π√163) is an integer—this was an April Fool's hoax.
• Don't confuse: "very close to an integer" with "is an integer"; these are transcendental numbers, not integers.

🔢 Fermat numbers and false conjectures

Example 5.1.4 discusses Fermat numbers F_a = 2^(2^a) + 1.

• The first five Fermat numbers are all prime: F₀ = 3, F₁ = 5, F₂ = 17, F₃ = 257, F₄ = 65537.
• Fermat conjectured F_a is always prime based on these cases.
• Counterexample: In 1732, Euler showed F₅ = 4294967297 = 641 · 6700417, which is composite.
• Currently no one knows how many Fermat numbers are prime; even with modern computing, no more Fermat primes have been found beyond F₄.
• Lesson: Even a pattern holding for the first five cases can fail at the sixth.

🎲 Pseudo-primes and Carmichael numbers

Example 5.1.5 examines Fermat's little theorem and its converse.

Fermat's little theorem:

If p is prime, then p divides a^(p−1) − 1 for all a such that gcd(a, p) = 1.

• Example: p = 5 divides a⁴ − 1 for a = 1, 2, 3, 4.
• Question: Is the converse true? If n divides a^(n−1) − 1 for all a with gcd(a, n) = 1, must n be prime?
• Counterexample: n = 561 divides a⁵⁶⁰ − 1 for all such a, but 561 = 3 · 11 · 17 is composite.
• 561 is a Carmichael number, also called a pseudo-prime (fake prime).
• Don't confuse: satisfying a property of primes with actually being prime.

🏛️ Why proofs are valuable

🏛️ Beyond verification

The excerpt emphasizes that proofs do more than just confirm truth:

• Show without doubt: proofs establish that a statement is true in all cases, not just the cases we can check.
• Explain why: proofs reveal the reasons behind a statement, not just that it holds.
• Beauty and power: many mathematicians find proofs beautiful and have favorite proofs that illustrate core ideas in their fields.
• Foundation vs language: some view proofs as the solid base needed to build mathematics; others see them as a living language connecting ancient and modern thought.

🏆 Famous proof achievements

Problem	Key facts	Outcome
Four Color Theorem (Example 5.1.3)	Any map can be colored with four colors so no adjacent countries share a color	Proven in 1976 by Appel and Haken using a computer to exhaustively check cases after categorizing possible maps
Fermat's Last Theorem (Example 5.1.6)	X^n + Y^n = Z^n has no integer solutions for n ≥ 3 unless XYZ = 0	Conjectured in 1637; motivated centuries of work in algebraic number theory and arithmetic geometry; proven in 1994 by Wiles (building on work of Frey, Ribet, Serre, Taniyama, Shimura); Wiles received the Abel Prize (~$700,000)

💻 Computers in proofs

• The Four Color Theorem proof used a computer to check many cases after mathematical reasoning reduced the problem to a finite (but large) set of cases.
• Computers can make proofs much faster when there are many cases to check.
• Fermat numbers remain useful because they can have very large prime factors; in 2020, a megaprime factor of F₅₅₂₃₈₅₈ was found as part of the PrimeGrid collaboration.

🎯 What this chapter will cover

🎯 Learning to write proofs

• Writing proofs is not easy; students sometimes feel it is like drill exercises.
• The chapter will explain proof techniques often useful in combinatorics:
  • Induction
  • Counting in two ways
  • Bijective proofs
  • Proof by contradiction
  • The pigeonhole principle
• Goal: by the end, writing a proof should seem more manageable; maybe you will have a favorite proof too.

🎯 Why this matters for combinatorics

• The chapter focuses on proof techniques specifically useful in combinatorics.
• Proofs are needed to show statements are true without doubt, especially when patterns in data can be misleading.
• The examples (divisors, Fermat numbers, pseudo-primes) all illustrate combinatorial or number-theoretic phenomena where intuition from small cases fails.

🧭 Overview

🧠 One-sentence thesis

Mathematical induction proves that a property holds for all natural numbers by establishing a base case and showing that each true case implies the next, like dominoes falling in sequence.

📌 Key points (3–5)

• The domino analogy: induction works by knocking over the first domino (base case) and ensuring each falling domino knocks over the next (inductive step), causing all infinitely many to fall.
• Two required components: you must prove both (1) the property is true for n = 0 (or some starting value), and (2) if true for an arbitrary k, then true for k+1.
• Common confusion: assuming what you want to prove—you cannot start by writing "A = B" and simplify; instead, assume P_k is true and use it to prove P_k+1.
• When statements fail for small n: if a property is false for small values but true from some c onward, modify the base case to start at n = c instead of n = 0.
• Why it works: the inductive step applies to an arbitrary k, so combined with the base case, it guarantees the property holds for every natural number.

🎯 The inductive principle

🎯 What induction proves

The inductive principle: If (1) property P_n is true for n = 0, and (2) whenever P_k is true for an arbitrary integer k ≥ 1, then P_(k+1) is also true, then P_n is true for all natural numbers n.

• The excerpt defines the natural numbers as N = {0, 1, 2, 3, ...}.
• A property P_n can be a formula or fact that might be true or false for each n.
• Example from the excerpt: P_n: "1 + 2 + ... + n = n(n+1)/2" is true for all n ≥ 1.
• Counter-example: P_n: "The Fermat number F_n is prime" is true for n = 1, 2, 3, 4 but false for n = 5, so it cannot be proven by any method.

🪜 The three-step method

Step 1: Base case

• Prove that P_n is true for n = 0 (or whatever starting value is appropriate).
• The excerpt notes this is usually an easy computation.

Step 2: Inductive hypothesis

• Let k ∈ N be arbitrary and assume P_k is true for this specific value k.
• The statement P_k is called the inductive hypothesis.
• Choosing k to be arbitrary means showing that no matter what domino falls, it will always knock over the next one.

Step 3: Inductive step

• Using the assumption that P_k is true, prove that P_(k+1) is true.
• Steps 2 and 3 together prove the "if...then" statement: to prove such a statement in mathematics, assume the hypothesis is true and then prove the conclusion holds under that assumption.

🔄 Aesthetic variation

• The excerpt mentions you can replace the inductive step with: if P_(k-1) is true for an arbitrary integer k ≥ 1, then P_k is true.
• In other words, in step 2 let k ≥ 1 (rather than k ≥ 0) and assume P_(k-1), then in step 3 prove P_k.
• This is a purely aesthetic choice but sometimes makes the algebra simpler.

📐 Worked examples

📐 Sum of first n odd integers

Property: P_n: 1 + 3 + 5 + ... + (2n - 1) = n² for all integers n ≥ 1.

Base case: For n = 1, note that 1 = 1².

Inductive step: Let k ≥ 1 be arbitrary and suppose P_k is true, meaning 1 + 3 + 5 + ... + (2k - 1) = k². Adding 2k + 1 to both sides gives:

• 1 + 3 + 5 + ... + (2k - 1) + (2k + 1) = k² + 2k + 1 = (k + 1)²
• This shows P_(k+1) is true.

Hence by induction, P_n is true for all n ≥ 1.

📐 Sum of consecutive integers

Property: P_n: 1 + 2 + ... + n = n(n+1)/2 for all integers n ≥ 1.

The excerpt references this as Lemma 1.2.1 and states it can be proven by induction (left as an exercise).

📐 Sum of powers of 2

Property: 1 + 2 + 4 + 8 + ... + 2^(n-1) = 2^n - 1 for all integers n ≥ 1.

The excerpt states this can be proven by induction (left as an exercise).

📐 Factorial inequality

Property: Prove that n! ≥ 2^n for all n ≥ 4.

• The excerpt notes the statement is false when n = 1, 2, 3 but true when n = 4.
• Base case: The identity is true when n = 4 because 24 = 4! ≥ 2⁴ = 16.
• Inductive step: Let k ≥ 4 be arbitrary and suppose k! ≥ 2^k. Then:
  • (k+1)! = (k+1) · k! ≥ (k+1) · 2^k ≥ 5 · 2^k ≥ 2^(k+1)
  • This shows P_(k+1) is true.

Hence by induction, P_k is true for all n ≥ 4.

🔧 Modified base cases

🔧 When to start at c instead of 0

Sometimes a statement P_n is false for small values of n. If there is a positive integer c for which P_c is true and the inductive step works, then we can modify the method of induction to prove that statement P_n is true for all n ≥ c (instead of n ≥ 0).

How to modify:

• Change the base case to prove P_c rather than prove P_0.
• Change the assumption on the inductive step to choose k ≥ c and assume P_k.

Example: The factorial inequality n! ≥ 2^n is false for n = 1, 2, 3, so the base case starts at n = 4 and the inductive step assumes k ≥ 4.

⚠️ Common mistakes

⚠️ Assuming what you want to prove

One common mistake that people make in writing proofs is to assume A = B and then simplify. Warning: this does not always work!

False example from the excerpt:

• Claim: 3 = 5.
• "Proof": Note 3 = 5 implies 3·0 = 5·0 implies 0 = 0 which is true.
• Why it's wrong: You cannot start by assuming the conclusion and work backward; the logic is reversed.

⚠️ Inductive step failing for small k

False example: All horses are the same color

The excerpt presents a "proof" by induction:

• Base case: In any herd of 1 horse, all horses in that herd are the same color. ✓
• Inductive step: Let k ≥ 1 be arbitrary and suppose all horses in a herd of size k are the same color. Consider a herd of k+1 horses:
  • Removing the last horse, the first horse is the same color as the horses in the middle.
  • Removing the first horse, the last horse is the same color as the horses in the middle.
  • Hence all k+1 horses are the same color.

Where it goes wrong: The inductive step fails when going from k = 1 to k = 2, even though it works for all larger values of k. When k = 1, the middle herd is empty, so the argument breaks down. This is why the bound on k in the declaration "Let k ≥ 1 be arbitrary" is so important.

Don't confuse: An inductive step that works for most values of k is not enough; it must work for every k starting from the base case.

🧮 Two-variable induction

🧮 Hockey Stick Identity

Property: For positive integer n and non-negative integer m,

• (n choose 0) + (n+1 choose 1) + (n+2 choose 2) + ... + (n+m choose m) = (n+m+1 choose m)

Key point: There are two variables n and m in the formula, so the excerpt specifies that induction proceeds on m for a fixed n.

Base case: The identity is true when m = 0 because (n choose 0) = 1 = (n+0+1 choose 0).

Inductive step: Let m ∈ N be arbitrary and assume:

• (n choose 0) + (n+1 choose 1) + (n+2 choose 2) + ... + (n+m choose m) = (n+m+1 choose m)

Must show:

• (n choose 0) + (n+1 choose 1) + (n+2 choose 2) + ... + (n+m choose m) + (n+m+1 choose m+1) = (n+m+2 choose m+1)

By substituting the inductive hypothesis, the left-hand side simplifies to:

• (n+m+1 choose m) + (n+m+1 choose m+1)

which equals (n+m+2 choose m+1) by Pascal's recurrence (Theorem 4.1.1).

Hence the statement is true for all non-negative integers m by induction.

🔍 Strong induction

🔍 Variation of the method

There is a variation of the inductive process, called Strong induction. In the inductive step of this variant, we suppose that P_k is true for all 1 ≤ k ≤ n, and show that P_k is true when k = n+1.

• In regular induction, you assume P_k is true for one specific k and prove P_(k+1).
• In strong induction, you assume P_k is true for all values from 1 up to n, and prove P_(n+1).
• This gives you more to work with in the inductive step.

🧩 Counting in two ways

🧩 The principle

Counting in Two Ways: If a set has n elements and it also has m elements then n = m.

• The excerpt calls this a "trivial principle" but notes it is incredibly useful in proving algebraic identities in combinatorics.
• How to use it: To prove F = G where F and G are both formulas, find a collection of objects such that both F and G are the answer to "how many are there in the collection?"

🧩 Multiplication as repeated addition

Property: a · b = b + b + ... + b (a summands), where a and b are positive integers.

• Right-hand side: By repeated application of the Addition Principle, b + b + ... + b counts the number of students in a school that has a classes with b students each.
• Left-hand side: By the Multiplication Principle, a · b counts the number of ways to first walk into one of the a classrooms and then single out one of the b students in that class.
• Thus a · b is also equal to the total number of students, so a · b = b + b + ... + b.

🧩 Pascal's recurrence combinatorially

Property: (n+1 choose k) = (n choose k-1) + (n choose k)

Combinatorial proof: Find the number of ways of choosing k people from a classroom that has n students and 1 professor.

• Since there are n+1 people in the room, the answer is (n+1 choose k).
• On the other hand, separate the ways of choosing the group into:
  • (i) those that do not contain the teacher: (n choose k) choices
  • (ii) those that do contain the teacher: (n choose k-1) choices (choose k-1 from the n students)
• By the Addition Principle, the total is (n choose k-1) + (n choose k).

Hence (n+1 choose k) = (n choose k-1) + (n choose k).

🧭 Overview

🧠 One-sentence thesis

Counting in two ways proves algebraic identities by showing that two different formulas both count the same collection of objects, so they must be equal.

📌 Key points (3–5)

• The principle: if a set has n elements and also has m elements, then n = m—sounds trivial but is powerful for proving combinatorial identities.
• How to use it: to prove F = G, find a collection of objects such that both F and G answer "how many are there in the collection?"
• Proof strategy: count the same set in two different ways—one way gives the left-hand side of the identity, the other way gives the right-hand side.
• Common confusion: the principle is not about comparing two different sets; it's about counting the same set using two different methods.
• Why it matters: provides combinatorial (intuitive, story-based) proofs for algebraic formulas involving binomial coefficients and sums.

🧩 The core principle

🧩 What "counting in two ways" means

Lemma (Counting in Two Ways): If a set has n elements and it also has m elements, then n = m.

• This sounds obvious, but it is the foundation for proving many combinatorial identities.
• The key insight: if you can count the same collection using two different methods, the two answers must be equal.
• Example: if you count students by adding up each classroom's size, or by multiplying number of classrooms times students per classroom, you get the same total.

🔍 How to apply the principle

• Goal: prove that formula F equals formula G.
• Method:
  • Find a concrete collection of objects (e.g., students, subsets, choices).
  • Show that F counts this collection one way.
  • Show that G counts the same collection a different way.
  • Conclude F = G.
• Don't confuse: you are not counting two different sets and comparing them; you are counting one set in two ways.

🎯 Simple examples

🎯 Proving multiplication equals repeated addition

The excerpt proves that a · b = b + b + ⋯ + b (with a summands).

• The collection: all students in a school with a classes, each having b students.
• First way (right-hand side): by the Addition Principle, b + b + ⋯ + b (a times) counts the total students.
• Second way (left-hand side): by the Multiplication Principle, a · b counts the number of ways to walk into one of a classrooms and single out one of b students—this also equals the total number of students.
• Conclusion: both formulas count the same thing, so they are equal.

🎯 Proving Pascal's recurrence

The excerpt proves that (n+1 choose k) = (n choose k−1) + (n choose k).

• The collection: ways to choose k people from a room with n students and 1 professor (total n+1 people).
• First way (left-hand side): (n+1 choose k) counts all such choices.
• Second way (right-hand side): separate into two cases:
  • (i) Choices that do not contain the professor: (n choose k) ways.
  • (ii) Choices that do contain the professor: (n choose k−1) ways (choose only k−1 students).
  • By the Addition Principle, total is (n choose k−1) + (n choose k).
• Conclusion: both count the same collection, so the identity holds.

🧮 Advanced examples

🧮 Sum of squares formula

The excerpt proves that (n choose 0)² + (n choose 1)² + ⋯ + (n choose n)² = (2n choose n).

Proof:

• The collection: ways to choose n soccer players out of 2n total players.
• First way (left-hand side): (2n choose n).
• Second way (right-hand side):
  • Divide the 2n players into two groups: n with gold jerseys, n with green jerseys.
  • To choose n players total, choose k from gold and n−k from green.
  • For a fixed k, there are (n choose k) · (n choose n−k) ways.
  • Since (n choose n−k) = (n choose k), this is (n choose k)².
  • Sum over all possible k from 0 to n: Σ (n choose k)² = (2n choose n).
• Conclusion: both formulas count the same collection.

🧮 Vandermonde's identity

The excerpt proves that (n+m choose ℓ) = Σ (from k=0 to ℓ) (n choose k)(m choose ℓ−k).

Proof:

• The collection: ways to choose ℓ snacks from n granola bars and m bags of trail mix (total n+m snacks).
• First way (left-hand side): (n+m choose ℓ).
• Second way (right-hand side):
  • Let D_k be the number of ways to choose exactly k granola bars and ℓ−k bags of trail mix.
  • D_k = (n choose k)(m choose ℓ−k).
  • Sum over all possible k from 0 to ℓ: Σ D_k = Σ (n choose k)(m choose ℓ−k).
• Conclusion: both count the same collection.
• Special case: substitute m = n = ℓ in Vandermonde's identity to get (2n choose n) = Σ (n choose i)², which is another proof of the sum of squares formula.

🔧 Convention and notation

🔧 Handling out-of-range binomial coefficients

• In the summation for Vandermonde's identity, the excerpt sets (n choose k) = 0 if k > n.
• Similarly, (m choose ℓ−k) = 0 if ℓ−k > m.
• This convention ensures the sum is well-defined even when some terms would otherwise be undefined.

🔧 Exercises hint

The excerpt lists several identities to prove by counting in two ways:

• (n choose k) = (n choose n−k): count subsets of size k vs. their complements of size n−k.
• (n choose 0) + (n choose 1) + ⋯ + (n choose n) = 2^n: count all subsets of an n-element set.
• (n choose k) · k = n · (n−1 choose k−1): count ways to choose a committee and designate a leader.
• Hockey Stick identity: count paths or cumulative choices.

🔗 Connection to bijections

🔗 Preview of bijective proofs

• The excerpt mentions that counting in two ways is not always enough; sometimes you need a bijection to compare the sizes of two different sets.
• Example question: how many ways can you choose 4 numbers from {1, …, 100} so that none are consecutive?
• This requires pairing elements of one set (4-element subsets with no consecutive numbers) with elements of another set (some easier-to-count collection) via a bijection.
• Don't confuse: counting in two ways counts one set in two ways; bijective proofs establish a one-to-one correspondence between two different sets to show they have the same size.

🧭 Overview

🧠 One-sentence thesis

Bijective proofs establish that two sets have the same size by constructing a one-to-one correspondence between their elements, enabling solutions to counting problems that are difficult to solve directly.

📌 Key points (3–5)

• What a bijection is: a map between two sets that is both 1-to-1 (no element is hit twice) and onto (every element is hit), creating a perfect pairing.
• Why bijections matter for counting: if there is a bijection between sets A and B, then A and B have the same number of elements, even if counting one set directly is hard.
• How to construct a bijection: define a map f from A to B and show it has an inverse map g from B to A that reverses the process.
• Common confusion: bijections differ from "counting in two ways"—bijections compare two different sets, while counting in two ways counts one set using two methods.
• Practical technique: transform a hard counting problem into an easier one by finding a bijection to a set whose size is already known.

🔗 What bijections are

🔗 Formal definition

Bijection: A map f : A → B that is both 1-to-1 and onto.

• 1-to-1 (injective): whenever a₁ ≠ a₂, then f(a₁) ≠ f(a₂); equivalently, if f(a₁) = f(a₂), then a₁ = a₂.
  • Intuition: no element of B is hit twice.
• Onto (surjective): for every b ∈ B, there exists an a ∈ A such that f(a) = b.
  • Intuition: the map hits every element of B; the range of f is all of B.
• A bijection creates a perfect pairing or matching between elements of A and elements of B.

🔄 Inverse maps

A bijection f : A → B has an inverse map g : B → A if and only if g ∘ f is the identity on A and f ∘ g is the identity on B.

• How to define the inverse: Given b ∈ B, since f is onto, there exists a unique a ∈ A (unique because f is 1-to-1) such that f(a) = b; define g(b) = a.
• Why this works: (g ∘ f)(a) = g(f(a)) = g(b) = a and (f ∘ g)(b) = f(g(b)) = f(a) = b.
• Conversely: if g is an inverse of f, then f is a bijection:
  • To show 1-to-1: if f(a₁) = f(a₂), then g(f(a₁)) = g(f(a₂)), so a₁ = a₂.
  • To show onto: for any b ∈ B, let a = g(b); then f(a) = f(g(b)) = b.

Example: Let A be positive integers ending in 3, B be positive integers ending in 7. The map f(a) = a + 4 is a bijection with inverse g(b) = b − 4.

Example: Let A be all integers, B be all multiples of 7. The map f(a) = 7a is a bijection with inverse g(b) = b/7.

🧮 The bijective proof principle

🧮 Core counting rule

Lemma (Bijective proof): If A has m elements, B has n elements, and there is a bijection f : A → B, then m = n.

• This sounds simple but can be tricky to execute in practice.
• Strategy: to prove two quantities are equal, construct two sets A and B whose sizes represent those quantities, then find a bijection between them.
• Don't confuse with "counting in two ways": bijections compare two different sets; counting in two ways counts one set using two different methods.

🎯 Classic examples

🎯 Subsets of complementary size

Problem: Show that the number of k-element subsets of {1, …, n} equals the number of (n − k)-element subsets.

• Let A be the set of k-element subsets of Ω = {1, …, n}.
• Let B be the set of (n − k)-element subsets of Ω.
• Bijection: map each subset S ∈ A to its complement Ω − S ∈ B.
• Since every subset has exactly one complement, this is a bijection.
• Conclusion: the number of k-element subsets equals the number of (n − k)-element subsets, i.e., (n choose k) = (n choose n−k).

🎯 Even-sized vs odd-sized subsets

Problem (Proposition 5.4.9): For Ω = {1, …, n}, show that the number of even-sized subsets equals the number of odd-sized subsets.

This proves the alternating sum identity: (n choose 0) − (n choose 1) + (n choose 2) − … ± (n choose n) = 0, which can be rewritten as (n choose 0) + (n choose 2) + (n choose 4) + … = (n choose 1) + (n choose 3) + (n choose 5) + …

Case 1: n is odd

• If S has size k, then Ω − S has size n − k.
• If k is even, then n − k is odd (since n is odd), and vice versa.
• Bijection: map each even-sized subset S to its complement Ω − S, which is odd-sized.
• Example: For Ω = {1, 2, 3}, pair ∅ ↔ {1,2,3}, {1} ↔ {2,3}, {2} ↔ {1,3}, {3} ↔ {1,2}.

Case 2: n is even

• The complement trick doesn't work (even − even = even).
• New bijection: Pick a distinguished element, say 1 ∈ Ω. Group subsets into pairs where the two subsets differ only in whether they contain 1.
• Each pair contains one even-sized subset and one odd-sized subset.
• Example: For Ω = {1,2,3,4}, pair {1,2} ↔ {2}, {1,3} ↔ {3}, {1,2,3} ↔ {2,3}, etc.
• Every subset is in exactly one pair, so the counts are equal.

Conclusion: Since there are 2ⁿ subsets total, the number of even-sized subsets is 2ⁿ / 2 = 2ⁿ⁻¹.

🍬 Multisets and the M&M's proof

🍬 Choosing with repetition allowed

Problem (Proposition 5.4.12): How many ways can you choose a set of size k from a set of size n with repeats allowed?

• Answer: (n + k − 1 choose k).
• This reproves the "sticks and stones" theorem (Theorem 3.5.6) using a bijection.

🍬 The bijection construction

Set A: cups filled with k M&M's, where there are n different possible colors.

• Size of A = number of ways to choose k items from n types with repetition.

Set B: binary sequences with k zeroes and (n − 1) ones.

• Total length: n + k − 1.
• Size of B = (n + k − 1 choose k), since we choose which k positions contain the 0's.

Map f : A → B (M&M's to binary sequence):

1. Spill the k M&M's onto a plate and sort them by color (using a fixed color ordering, e.g., rainbow order).
2. The M&M's are now separated into n sections (some may be empty).
3. Place a toothpick between each section, using (n − 1) toothpicks total.
4. Put on color-filter goggles so all M&M's look grey.
5. Write a 0 for each grey M&M and a 1 for each toothpick.
6. Result: a sequence of k zeroes and (n − 1) ones.

Inverse map g : B → A (binary sequence to M&M's):

1. Interpret the sequence as k grey M&M's separated by (n − 1) toothpicks.
2. Color the M&M's according to the specified ordering of colors.
3. Put them into a cup.

Since these processes reverse each other, f and g are inverses, so f is a bijection.

🔢 Non-consecutive number selection

🔢 The motivating problem

Question 5.4.1: How many ways can you choose 4 numbers from {1, …, 100} so that none of them are consecutive?

• We know there are (100 choose 4) ways to choose 4 numbers, but the "non-consecutive" condition is hard to handle directly.
• Strategy: transform the problem into an easier one via bijection.

🔢 The transformation

Set A: 4-tuples (a₁, a₂, a₃, a₄) with 1 ≤ a₁ < a₂ − 1 < a₃ − 2 < a₄ − 3 ≤ 97.

• This encodes the non-consecutive condition: a₂ > a₁ + 1, a₃ > a₂ + 1, a₄ > a₃ + 1.

Set B: 4-tuples (b₁, b₂, b₃, b₄) with 1 ≤ b₁ < b₂ < b₃ < b₄ ≤ 97.

• This is just choosing 4 distinct numbers from {1, …, 97}.
• Size of B = (97 choose 4).

Bijection f : A → B:

• Define f(a₁, a₂, a₃, a₄) = (a₁, a₂ − 1, a₃ − 2, a₄ − 3).
• Let b₁ = a₁, b₂ = a₂ − 1, b₃ = a₃ − 2, b₄ = a₄ − 3.
• The condition a₂ > a₁ + 1 becomes b₂ + 1 > b₁ + 1, i.e., b₂ > b₁.
• Similarly, a₃ > a₂ + 1 becomes b₃ > b₂, and a₄ > a₃ + 1 becomes b₄ > b₃.
• Also, a₄ ≤ 100 becomes b₄ ≤ 97.

Inverse g : B → A:

• Define g(b₁, b₂, b₃, b₄) = (b₁, b₂ + 1, b₃ + 2, b₄ + 3).
• Since f and g reverse each other, f is a bijection.

Answer: The number of ways to choose 4 non-consecutive numbers from {1, …, 100} is (97 choose 4).

🔢 Why this works

• The shifts (−1, −2, −3) "absorb" the gaps required by the non-consecutive condition.
• The transformed problem (choosing 4 distinct numbers from a smaller range) is straightforward to count.
• Don't confuse: the original problem has a₁ < a₂ < a₃ < a₄ with gaps; the transformed problem has b₁ < b₂ < b₃ < b₄ with no gap requirement, but from a smaller set.

🧭 Overview

🧠 One-sentence thesis

Proof by contradiction establishes a claim by assuming its opposite and deriving an impossible conclusion, which forces the original claim to be true.

📌 Key points (3–5)

• Core strategy: assume the negation of what you want to prove, then show this assumption leads to something definitely false.
• When to use it: especially useful when no direct formula or construction is available (e.g., proving infinitely many primes exist).
• Common confusion: don't overuse—a direct proof is better when available; contradiction is fun but not always necessary.
• Number theory applications: the excerpt demonstrates irrationality proofs (√3) and Euclid's proof of infinitely many primes.

🔄 The contradiction strategy

🔄 How the method works

Proof by contradiction: assume the opposite (negation) of what you want to show; then show that this implies something that is definitely false; this means that the initial assumption must have been false.

• You start by supposing the claim is not true.
• You follow logical steps from that assumption.
• You arrive at a statement that contradicts known facts or your own assumptions.
• Because the chain of reasoning was valid, the starting assumption must be wrong.
• Therefore, the original claim must be true.

✈️ Everyday example

Claim: If I want to catch my 10 am flight from Denver, I need to leave the house before 7 am.

Proof sketch:

• Assume I leave at 7 am or later.
• Then I arrive at the parking lot by 8:15, at the airport by 8:30, finish bag check by 8:45, clear security by 9:00, take the train by 9:15, and reach the gate at 9:45.
• But the doors close before 9:45—this is too late.
• So the assumption (leaving at 7 or later) leads to missing the flight, which contradicts the goal.
• Therefore, I must leave before 7 am.

🔢 Number theory proofs

🔢 Proving √3 is irrational

Claim: √3 is not a fraction.

Proof outline:

• Assume √3 is rational, so √3 = a/b where a and b are integers, b ≠ 0, and gcd(a, b) = 1 (the fraction is in simplest form).
• Squaring both sides: 3 = a²/b², so 3b² = a².
• This means 3 divides a². Since 3 is prime, 3 must divide a (by unique prime factorization).
• Write a = 3a₁ for some integer a₁.
• Substitute: 3b² = (3a₁)² = 9a₁², so b² = 3a₁².
• Then 3 divides b² and therefore 3 divides b.
• But now both a and b are divisible by 3, so gcd(a, b) is a multiple of 3—contradicting gcd(a, b) = 1.
• Therefore, √3 cannot be a fraction.

Key insight: The contradiction arises because assuming √3 is rational forces both numerator and denominator to share a common factor, violating the simplest-form assumption.

∞ Euclid's proof of infinitely many primes

Claim: There are infinitely many prime numbers.

Historical note: Euclid discovered this proof around 300 B.C. Contradiction is used because no one has found a formula to generate all primes directly.

Proof outline:

• Assume there are only finitely many primes. Let r be the total number of primes, and let S = {p₁, …, pᵣ} be the complete set of all primes.
• Consider the number N = 1 + (product of all primes in S) = 1 + p₁ · p₂ · … · pᵣ.
• N must be divisible by at least one prime q (every positive integer has a unique prime factorization).
• But N is not divisible by any prime in S, because N ≡ 1 mod pᵢ for each i (dividing N by any pᵢ leaves remainder 1).
• So q is a new prime not in S.
• This contradicts the assumption that S contains all primes.
• Therefore, there must be infinitely many primes.

Why contradiction is needed: There is no known algorithm to list all primes, so we cannot prove the claim directly by construction.

⚠️ When to use contradiction

⚠️ Appropriate use cases

• When no direct construction or formula exists (e.g., proving infinitely many primes).
• When the negation of the claim is easier to work with logically.
• When you need to show existence without being able to exhibit the object directly (the excerpt mentions "combinatorial existence proofs").

⚠️ Don't overuse

Sometimes proofs by contradiction are so fun that people use them more often than needed. It is better to use a direct proof when you can.

• Contradiction can be elegant, but it is not always the simplest approach.
• If you can prove something directly, that is usually clearer and more informative.
• Example: Don't assume "not P" and derive P when you can just prove P from the start.

🧭 Overview

🧠 One-sentence thesis

The pigeonhole principle proves that when more objects are distributed into fewer containers than objects, at least one container must hold multiple objects, enabling proofs of grouping and existence in combinatorics.

📌 Key points (3–5)

• Core mechanism: If you place more than n objects into n boxes, at least one box must contain 2 or more objects (Version 1).
• Generalized version: If you place more than k·n objects into n boxes, at least one box must contain k+1 or more objects (Version 2).
• Proof strategy: Uses contradiction—assume each box has at most the threshold number of objects, then show the total would be too small.
• Common confusion: The principle says nothing about how many boxes are empty; all objects could be in one box.
• Problem-solving approach: Identify k+1 (your goal), define n boxes cleverly so objects in the same box satisfy your goal, then verify object count exceeds k·n.

🎯 What the principle states

🎯 Version 1: Basic form

Pigeonhole principle Version 1: If we place more than n objects into n boxes, then some box contains 2 or more objects.

• Plain language: More items than containers guarantees at least one container holds multiple items.
• Why it works: If each box held at most 1 object, the total would be at most n objects—contradicting that we have more than n.
• Example: 11 numbers chosen from 1 to 100, divided into 10 boxes (1–10, 11–20, ..., 91–100). Since 11 > 10, two numbers must share a box, so their difference is less than 10.

🎯 Version 2: Generalized form

Pigeonhole principle Version 2: Suppose k is a non-negative integer. If we place more than k·n objects into n boxes, then some box contains at least k+1 objects.

• Relationship to Version 1: Substitute k=1 into Version 2 to recover Version 1.
• Why it works: If each box held at most k objects, the total would be at most k·n objects—contradicting that we have more than k·n.
• Example: Colorado population 5,612,000 people, n=1,000,000 boxes (one per possible hair count 0 to 999,999), k=5. Since 5,612,000 > 5·1,000,000, at least one box contains 6 or more people with the same hair count.

⚠️ What the principle does NOT say

• Empty boxes: The principle says nothing about how many boxes are empty.
• Distribution: It is possible that all objects are stuffed into the same box.
• Don't confuse: The principle guarantees at least one box meets the threshold; it does not describe the distribution across all boxes.

🔧 How to apply the principle

🔧 Three-step problem-solving strategy

The excerpt recommends:

1. Identify k+1: Determine your goal (e.g., "at least 6 people" means k+1=6, so k=5).
2. Define n boxes: Design boxes so that if enough objects land in a single box, you accomplish the problem's goal.
3. Verify object count > k·n: Confirm the number of objects exceeds k·n to trigger the principle.

📐 Geometric example: Triangle target

Problem: An equilateral triangle target with side-length 2 meters is hit by 5 arrows. Prove two arrows strike within 1 meter of each other.

Solution:

• Divide the triangle into n=4 smaller equilateral triangles, each with side-length 1 meter (the "boxes").
• The 5 arrow striking points are the "objects."
• Since 5 > 4, by Version 1, at least one smaller triangle contains at least 2 striking points.
• The distance between any two points in the same smaller triangle is at most 1 meter (the side length).

🤝 Handshake example: Hidden constraint

Problem: At a party with n guests (n ≥ 2), some pairs shake hands (no one shakes their own hand). Prove at least two guests shook hands with the same number of people.

Solution:

• Each guest shook between 0 and n−1 hands.
• Label boxes 0 through n−1; place a person in box i if they shook exactly i hands.
• Key insight: Either box 0 or box n−1 must be empty:
  • If someone shook 0 hands, then no one shook all n−1 hands.
  • If someone shook all n−1 hands, then no one shook 0 hands.
• So we assign n people to n−1 boxes (either 0 through n−2, or 1 through n−1).
• Since n > n−1, by Version 1, two people must be in the same box, so they shook the same number of hands.

🪑 Consecutive seating problems

🪑 Row of chairs

Problem: 17 people seated in a row of 20 chairs. Prove some consecutive set of 5 chairs are filled.

Solution:

• There are 3 empty chairs.
• These separate the row into n=4 "boxes" (far left, center left, center right, far right), each containing consecutively filled chairs.
• Let k+1=5, so k=4.
• Since 17 > 16 = 4·4 = k·n, by Version 2, some box contains at least 5 consecutively filled chairs.

🔄 Circle of chairs

Problem: 17 people seated in a circle of 20 chairs. Prove some consecutive set of 6 chairs are filled.

Solution:

• The 3 empty chairs partition the circle into n=3 "boxes" of consecutively filled chairs.
• Let k+1=6, so k=5.
• Since 17 > 15 = 5·3 = k·n, by Version 2, some box contains at least 6 consecutively filled chairs.

Don't confuse: Circle vs. row—the circle has fewer partitions (n=3 vs. n=4) because empty chairs in a circle create fewer segments, so you need a larger k to guarantee the same consecutive length.

🎨 Application to graph coloring (Ramsey theory)

🎨 Triangle in a colored complete graph

Problem: Consider the complete graph K₆ with 6 vertices (every pair connected by an edge). If you color the edges green and gold, show there must be either a green triangle or a gold triangle.

Solution:

• Pick one vertex v. It has 5 edges connecting it to the 5 other vertices.
• Let n=2 (for the 2 colors) and k=2.
• Since 5 > 2·2 = 4, by Version 2, at least 3 of these 5 edges have the same color (say green).
• These 3 green edges connect v to 3 other vertices w, x, y.
• Case analysis:
  • If any edge among w, x, y is green (e.g., w–x), then v, w, x form a green triangle.
  • If all three edges among w, x, y are gold, then w, x, y form a gold triangle.
• Either way, a monochromatic triangle exists.

Why this matters: This is an example of Ramsey theory, which studies unavoidable patterns in large structures.

🧭 Overview

🧠 One-sentence thesis

This problem set applies combinatorial reasoning, the pigeonhole principle, and modular arithmetic to prove identities, spatial constraints, and number-theoretic results like Fermat's Little Theorem.

📌 Key points (3–5)

• Combinatorial proofs: demonstrate identities by counting the same set in two different ways or by constructing bijections between sets.
• Pigeonhole principle applications: prove that certain configurations (consecutive filled chairs, nearby points, number differences) must exist when items are distributed into limited containers.
• Fermat's Little Theorem: states that a to the power p is congruent to a modulo p for any prime p and positive integer a; can be proven algebraically or combinatorially.
• Common confusion: pigeonhole problems in rows vs circles require different consecutive-set sizes because circles have no endpoints.
• Induction and well-ordering: these principles are closely related but differ in structure—induction builds upward step-by-step, while well-ordering guarantees a smallest element in any nonempty subset.

🧮 Combinatorial proof techniques

🧮 What combinatorial proofs are

Combinatorial proof: a demonstration that two expressions are equal by showing they count the same set in two different ways, or by constructing a bijection between two sets.

• Instead of algebraic manipulation, you interpret both sides of an equation as answers to counting questions.
• The excerpt asks for proofs of factorial formulas, binomial coefficient identities, and summation identities using these methods.

🔢 Factorial and power formulas

The problems ask to prove:

• n factorial equals n times (n−1) times (n−2) down to 3 times 2 times 1
• n to the power k equals the product of k copies of n
• The falling factorial n times (n−1) times ... times (n−k+1) equals n factorial divided by (n−k) factorial
• The binomial coefficient (n choose k) equals n factorial divided by k factorial times (n−k) factorial

How to approach: interpret each side as counting arrangements or selections.

• Example: n factorial counts the number of ways to arrange n distinct objects in a row; the product n·(n−1)·...·1 counts the same thing by choosing the first position (n choices), then the second (n−1 choices), etc.

🔗 Summation identities

The excerpt includes:

• Sum from k=0 to n of k times (n choose k) squared equals n times (2n−1 choose n−1)
• Sum from k=0 to n of (n choose k) times s to the power k times t to the power (n−k) equals (s+t) to the power n (binomial theorem with positive integers s, t, n)
• Sum from k=0 to n of 2 to the power k times (n choose k) equals 3 to the power n

How to approach: find a counting scenario where one side counts by cases and the other counts directly.

• Example: for the identity involving 2 to the power k times (n choose k), think of choosing a subset of n items and then assigning each chosen item one of two additional labels; the left side counts by subset size, the right side counts all possibilities at once (each of n items has 3 states: not chosen, chosen with label 1, chosen with label 2).

🕳️ Pigeonhole principle problems

🕳️ What the pigeonhole principle says

Pigeonhole principle: if more items are placed into fewer containers than there are items, at least one container must hold more than one item.

• The excerpt applies this to prove existence of number pairs with small differences, points within a certain distance, and consecutive filled chairs.

🔢 Number and point problems

Problem 8 (from earlier in the excerpt): if 17 numbers are chosen from 1 to 100, show that two of them have a difference less than 17.

• Divide the range 1–100 into intervals (pigeonholes); with 17 numbers and fewer than 17 intervals, two numbers must fall in the same interval, guaranteeing a small difference.

Problem 9: among any 10 points inside a unit square, there must exist 3 of them all within square root of 2 divided by 2 of each other.

• Partition the unit square into regions (pigeonholes); with 10 points and fewer regions, some region must contain at least 3 points, and the region's diameter ensures the distance bound.

🪑 Consecutive chair problems

The excerpt presents several variations:

Setup	Result	Key difference
94 people, 100 chairs in a row	Some consecutive 14 chairs are filled	Row has endpoints
94 people, 100 chairs in a circle	Some consecutive 16 chairs are filled	Circle has no endpoints; requires larger set
57 people, 64 chairs in a circle	At least 9 consecutively filled chairs	Fewer people relative to chairs
75 people, 80 chairs in a row	At least 13 consecutively filled chairs	More people relative to chairs

How to approach: divide the chairs into overlapping or non-overlapping groups (pigeonholes); count how many groups there are and how many must be "hit" by the people; deduce that some group must be densely filled.

Don't confuse: rows and circles require different arguments because in a circle, every position is equivalent (no first or last chair), which changes the pigeonhole structure.

🎯 Group difference problem

Problem 4: if 30 numbers are chosen from 1 to 82, show that there is a group of at least 5 of them such that the difference between any two numbers in this group is less than 12.

• Partition the range into intervals of width less than 12 (pigeonholes).
• With 30 numbers and fewer than 30 intervals, some interval must contain at least 5 numbers.
• All numbers in the same interval have pairwise differences less than 12.

🔁 Induction and well-ordering

🔁 Well-ordering principle

Problem 7a: look up the statement of the well-ordering principle and explain how it is similar to and different from induction.

• The excerpt does not provide the statement itself, but asks you to compare it to induction.
• The problem hints that well-ordering guarantees any nonempty subset of natural numbers has a least element.

🔁 Proving well-ordering from strong induction

Problem 7b: prove the well-ordering principle using only strong induction and the axiom that 0 is smaller than every positive integer.

• Define P(n) as "any subset of the natural numbers containing n has a least element."
• Base case: prove P(0)—if a subset contains 0, then 0 is the least element (by the axiom).
• Inductive step: assume P(0), P(1), ..., P(n) are all true; prove P(n+1).
  • If a subset contains n+1, either it contains some smaller number (in which case P(0)...P(n) guarantee a least element) or n+1 is the smallest.
• This shows that every subset has a least element, proving well-ordering.

Don't confuse: induction proves statements for all n by building upward; well-ordering guarantees a minimum exists in any nonempty set—they are logically equivalent but conceptually distinct.

🔐 Fermat's Little Theorem

🔐 Statement of the theorem

Fermat's Little Theorem: for any prime p and positive integer a, a to the power p is congruent to a modulo p.

• Equivalently: p divides (a to the power p minus a).
• Example: when p=7, then 7 divides (1 to the power 7 minus 1), 7 divides (2 to the power 7 minus 2), 7 divides (3 to the power 7 minus 3), etc.

🧪 Algebraic proof via binomial expansion

Problem 8: prove Fermat's Little Theorem using modular arithmetic and binomial coefficients.

• (a) Generalize a previous exercise to show: (x₁ + ... + xₐ) to the power p is congruent to (x₁ to the power p + ... + xₐ to the power p) modulo p for any a ≥ 2.
  • This uses the fact that binomial coefficients (p choose k) for 0 < k < p are divisible by p when p is prime.
• (b) Substitute x₁ = 1, ..., xₐ = 1 (so there are a copies of 1).
  • Left side: (1 + 1 + ... + 1) to the power p = a to the power p.
  • Right side: 1 to the power p + ... + 1 to the power p = a.
  • Therefore a to the power p ≡ a (mod p).

🎲 Combinatorial proof via sequences

Problem 9: prove Fermat's Little Theorem by counting sequences.

• (a) Let S be the set of sequences of length p with entries from the alphabet {1, ..., a} such that not all entries are the same.
  • Total sequences: a to the power p.
  • Sequences with all entries the same: a.
  • Therefore |S| = a to the power p minus a.
• (b) Define an equivalence relation: two sequences are equivalent if one is a rotation of the other.
  • Example: when p=5, (1,2,3,4,5) is equivalent to (2,3,4,5,1).
  • Each equivalence class has size p because there are p distinct rotations of any sequence (and they are all distinct because not all entries are the same).
  • Why p must be prime: if p were composite and a sequence had a repeating pattern with period dividing p, rotations would not all be distinct; primality ensures no such period exists.
• (c) By the division principle, the number of equivalence classes is |S| divided by p = (a to the power p minus a) divided by p.
  • This must be an integer, so p divides (a to the power p minus a).

Don't confuse: the algebraic proof uses properties of binomial coefficients; the combinatorial proof uses counting and symmetry—both are valid but use different reasoning.

🧭 Overview

🧠 One-sentence thesis

Pascal's triangle modulo 2 reveals a self-similar fractal pattern that exactly matches the structure of Sierpinski's triangle, demonstrating a deep connection between combinatorics and geometric fractals.

📌 Key points (3–5)

• What modulo 2 means for Pascal's triangle: replace all even binomial coefficients with 0 and all odd coefficients with 1 to reveal hidden patterns.
• Key structural property: rows that are powers of 2 (n = 2^m) contain only 1's at the first and last positions, with all middle entries being 0.
• Self-similarity pattern: the rows from n = 0 to n = 2^m - 1 repeat on both the left and right sides of rows n = 2^m to n = 2^(m+1) - 1, with a triangle of zeros in the middle.
• Common confusion: the pattern is not about the original binomial coefficient values but about their remainders when divided by 2 (odd vs even).
• Fractal connection: the 1's in Pascal's triangle mod 2 outline Sierpinski's triangle, a famous self-similar fractal shape.

🔢 Understanding Pascal's Triangle Modulo 2

🔢 Converting to modulo 2

If an integer a is even, then a ≡ 0 mod 2; if a is odd, then a ≡ 1 mod 2.

• The process: take each entry in Pascal's triangle and replace it with its remainder when divided by 2.
• Even numbers become 0, odd numbers become 1.
• Example from the excerpt:
  • Row n = 4: 1, 4, 6, 4, 1 becomes 1, 0, 0, 0, 1
  • Row n = 5: 1, 5, 10, 10, 5, 1 becomes 1, 1, 0, 0, 1, 1

🔍 Computing the next row

• You can determine the next row of Pascal's triangle mod 2 from the previous row without computing the original triangle.
• The excerpt asks students to discover this method through observation of patterns.

🎯 The Power-of-2 Row Property

🎯 Statement of the key proposition

Proposition 5.8.4: If n = 2^m is a power of 2, then the numbers in row n of Pascal's triangle mod 2 are all 0 except for the first and last entry.

• This means rows n = 1, 2, 4, 8, 16, 32, ... have the pattern: 1, 0, 0, ..., 0, 1
• After row n = 3 (which is 1, 1, 1, 1), the next all-1's row would be at the next power of 2.

🛠️ Three proof methods

The excerpt outlines three different approaches to prove this proposition:

Method	Key idea
Factorial/prime factorization	Show the numerator of (2^m choose k) has more factors of 2 than the denominator
Algebraic	Use induction to show (x + y)^(2^m) ≡ x^(2^m) + y^(2^m) mod 2
Bijective/combinatorial	Count subsets and show the count is even using complements and induction

🧮 The algebraic approach details

• Step 1: Show that (x + y)^2 ≡ x^2 + y^2 mod 2 (base case)
• Step 2: Use induction on m, assuming (x + y)^(2^(m-1)) ≡ x^(2^(m-1)) + y^(2^(m-1)) mod 2
• Step 3: Square both sides to get the result for 2^m
• The key insight: reducing coefficients (but not exponents) modulo 2 eliminates all middle terms

🔁 Self-Similarity and Repetition Patterns

🔁 The upside-down triangle of zeros

• Between rows n = 2^m and n = 2^(m+1) - 1, there is a triangular region of zeros in the middle.
• The excerpt uses a "ball-dropping" metaphor: balls placed at the edges of row n = 2^m can only land in certain positions, creating this zero region.

🪞 Left and right repetition

• The rows from n = 0 to n = 2^m - 1 repeat on both the left-hand side and right-hand side of rows n = 2^m to n = 2^(m+1) - 1.
• Mathematically: if 0 ≤ n ≤ 2^m - 1 and 0 ≤ k ≤ n, then (n choose k) ≡ (n + 2^m choose k) mod 2 and (n choose k) ≡ (n + 2^m choose k + 2^m) mod 2.

🧪 Proving the pattern

The excerpt shows that (x + y)^(n + 2^m) ≡ (x^(2^m) + y^(2^m)) · (x + y)^n mod 2.

• This algebraic identity explains both the upside-down triangle of zeros and the self-similar repetition.
• The factor (x^(2^m) + y^(2^m)) creates the splitting into left and right copies.

🔺 Connection to Sierpinski's Triangle

🔺 What is Sierpinski's triangle

Sierpinski's triangle is a famous example of a fractal, which is a shape that exhibits self-similarity.

• It is made up of three shrunken copies of itself, each half the total size, arranged in a triangle.
• Construction method 1: Start with a small triangle, copy it to lower left and lower right, repeat infinitely.
• Construction method 2: Start with a black equilateral triangle, divide into 4 smaller triangles, color the middle one white, repeat for each remaining black triangle.

🎨 The pattern match

• The pattern of 1's in Pascal's triangle mod 2 outlines a Sierpinski triangle shape.
• The self-similarity discovered in Pascal's triangle mod 2 (rows repeating on left and right) is exactly the same as the self-similarity in Sierpinski's triangle.
• Example activity: Print the first 16 rows of Pascal's triangle mod 2 and connect the 1's with lines in three triangular directions to see the Sierpinski pattern emerge.

🌀 Why this connection exists

• Both structures exhibit the same recursive pattern: a shape made of two copies of itself (left and right) with a gap in the middle.
• The algebraic structure of binomial coefficients modulo 2 naturally produces the geometric structure of the Sierpinski fractal.
• Don't confuse: Sierpinski's triangle is the infinite limit of the construction process, while any finite printout of Pascal's triangle mod 2 is only an approximation.

🧭 Overview

🧠 One-sentence thesis

The Fibonacci sequence, defined recursively as F_n = F_(n−1) + F_(n−2), appears in diverse counting problems—from rabbit populations to staircase climbing—and can be computed directly using a closed-form formula involving the golden ratio.

📌 Key points (3–5)

• What a recursive relation is: a formula that defines each term in a sequence using previous terms in that sequence.
• The Fibonacci recurrence: F_1 = 1, F_2 = 1, and for n ≥ 3, F_n = F_(n−1) + F_(n−2), producing the sequence 1, 1, 2, 3, 5, 8, 13, 21, ...
• Staircase problem connection: the number of ways to climb n stairs (taking 1 or 2 steps at a time) equals F_(n+1), proven by showing the same recurrence holds.
• Common confusion: the staircase sequence S_n and Fibonacci F_n follow the same recurrence but are shifted—S_n = F_(n+1), not S_n = F_n.
• Closed-form formula: F_n can be computed directly using the golden ratio α = (1 + √5)/2 without calculating all prior terms, though the formula surprisingly involves irrational numbers yet always yields integers.

🐇 The rabbit problem and Fibonacci definition

🐇 Original rabbit question

The Fibonacci numbers were first published in 1202 by Leonardo of Pisa in response to this question:

Suppose that rabbits mature in one month, that an adult pair of rabbits produces exactly one baby pair of rabbits each month, and that no rabbits ever die. Starting with one pair of baby rabbits in month 1, how many pairs of rabbits will there be in month 17?

• The answer is F_17 = 1597 pairs.
• The problem is completely unrealistic, but it led to a sequence that appears in nature (leaves on stems, pine cones, artichokes, pineapples, bee family trees) and applications in economics, logic, optics, and pseudo-random number generators.

📐 Formal definition

Recursive relation: A formula for a_n in terms of the previous values in the sequence a_1, a_2, ...

n-th Fibonacci number F_n: Set F_1 = 1 and F_2 = 1. For n ≥ 3, F_n = F_(n−1) + F_(n−2).

• Each term is the sum of the two preceding terms.
• The sequence begins: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, ...

🪜 The staircase climbing problem

🪜 The question

A staircase has n steps. How many ways can you go up the stairs if you can take one or two steps at a time and never go down?

• Let S_n be the number of ways to climb n steps.
• Example: for n = 4, there are 5 ways: (1,1,1,1), (1,1,2), (1,2,1), (2,1,1), (2,2).

🔍 Why the recurrence holds

The excerpt explains the logic for S_4 = S_3 + S_2:

• To climb 4 steps, you must start with either:
  • a size-1 step, leaving 3 stairs remaining → S_3 ways to proceed
  • a size-2 step, leaving 2 stairs remaining → S_2 ways to proceed
• By the addition principle, S_4 = S_3 + S_2.

More generally, for n ≥ 3:

• Every sequence of steps up n stairs starts with either a size-1 step (leaving n−1 stairs, so S_(n−1) ways) or a size-2 step (leaving n−2 stairs, so S_(n−2) ways).
• Therefore S_n = S_(n−1) + S_(n−2).

🔗 Relationship to Fibonacci numbers

Lemma 6.1.5: S_n = F_(n+1) for all n ≥ 1.

Proof sketch (using strong induction):

• Base cases: S_1 = 1 = F_2 and S_2 = 2 = F_3.
• Inductive step: Assume S_i = F_(i+1) for all 1 ≤ i ≤ n−1. Then:
  • S_n = S_(n−1) + S_(n−2) (by the recurrence for stairs)
  • = F_n + F_(n−1) (by the inductive hypothesis)
  • = F_(n+1) (by the Fibonacci recurrence).

Don't confuse: The staircase values S_n are Fibonacci numbers, but shifted left by one index compared to F_n. For example, S_5 = F_6 = 8.

n (stairs)	1	2	3	4	5
S_n	1	2	3	5	8
Corresponding F	F_2	F_3	F_4	F_5	F_6

🌟 The golden ratio and closed-form formula

🌟 What the golden ratio is

Golden ratio: α = (1 + √5)/2 ≈ 1.61803...

Conjugate: α̅ = (1 − √5)/2 ≈ −0.61803...

• Both α and α̅ are roots of the quadratic polynomial f(x) = x² − x − 1 (by the quadratic formula).
• The golden ratio is important in Greek architecture and natural objects like sea shells for the ratio of length to height.

📐 Closed-form formula for Fibonacci numbers

Theorem 6.1.6: For all n ≥ 0, F_n = (1/√5) × [ ((1+√5)/2)^n − ((1−√5)/2)^n ].

In words: F_n equals one over the square root of 5, times the difference between α to the n-th power and α̅ to the n-th power.

Why this is surprising:

• The right-hand side involves irrational numbers (√5) but always produces an integer.
• It is not initially clear why the Fibonacci numbers are connected to the golden ratio.

🔢 Practical consequence

• Drawback of recursion: computing F_100 requires computing F_1 through F_99 first.
• Closed-form advantage: the formula allows direct computation of F_n without computing all prior terms.
• Approximation: Because |α̅| < 1, powers of α̅ approach 0 as n grows large. Therefore F_n ≈ α^n for large n.

Note: The excerpt does not prove Theorem 6.1.6 in this chapter; it mentions that explanation appears in Section 6.2 (Example 6.2.14) and that induction proof will be revisited in Chapter 7.

🧮 Additional Fibonacci patterns (from exercises)

🧮 Sums and identities

The excerpt includes several patterns proven by induction:

Identity	Formula	Meaning
Sum of first n Fibonacci	F_1 + F_2 + ⋯ + F_n = F_(n+2) − 1	The sum telescopes to one less than F_(n+2)
Sum of odd-indexed	F_1 + F_3 + F_5 + ⋯ + F_(2n−1) = F_(2n)	Odd-indexed Fibonacci sum to an even-indexed one
Sum of squares	F₁² + F₂² + ⋯ + F_n² = F_n × F_(n+1)	Squares sum to product of consecutive terms

🎨 Other counting problems yielding Fibonacci

The exercises describe additional scenarios that produce Fibonacci numbers:

• Domino tiling: D_n = number of ways to cover an n-by-2 board with 2-by-1 dominoes (no overlap) equals F_(n+1).
• Subway seats: S_n = number of ways to fill n seats so no two people sit next to each other equals F_(n+2) for n ≥ 3.
• Rock painting: number of ways to paint n rocks silver or navy so no two silver rocks are adjacent equals F_(n+2).
• Subset without consecutive integers: the number of subsets of {1, 2, 3, ..., 9} containing no two consecutive integers is a Fibonacci number.

Common structure: All these problems involve "take it or leave it" choices with a constraint that prevents two consecutive selections, leading to the same recurrence S_n = S_(n−1) + S_(n−2).

🧭 Overview

🧠 One-sentence thesis

Linear recurrence relations can be solved systematically using the characteristic polynomial method to produce closed-form formulas that avoid the computational slowdown of recursive calculation.

📌 Key points (3–5)

• What a linear recurrence relation is: an equation defining each term as a weighted sum of previous terms (e.g., a_n = c₁·a_{n-1} + c₂·a_{n-2}).
• Why closed-form formulas matter: computers compute early entries quickly but slow down dramatically for later entries (e.g., the 100th term); closed formulas solve this.
• The characteristic polynomial method: convert the recurrence into a polynomial, find its roots, then express a_n as a sum of powers of those roots.
• Common confusion: the characteristic polynomial is not the recurrence relation itself—it is x^d - c₁·x^{d-1} - ... - c_d, where the signs flip and x replaces the index.
• Order matters: order 2 recurrences need 2 initial values and produce 2 roots; order d recurrences need d initial values and produce d roots.

🧩 Core definitions and examples

🧩 What is a linear recurrence relation?

Linear recurrence relation of order d: an equation of the form a_n = c₁·a_{n-1} + c₂·a_{n-2} + ... + c_d·a_{n-d}, where c₁, ..., c_d are constants and c_d ≠ 0.

• Each term is a linear combination of the previous d terms.
• The initial values are the first d terms: a₁, a₂, ..., a_d.
• Sometimes called "order d homogeneous linear recurrence with constant coefficients."

🪜 Example: Staircase climbing (order 3)

Problem: How many ways can you climb n stairs if you can take 1, 2, or 3 steps at a time?

• Let T_n = number of ways to climb n stairs.
• Recurrence: T_n = T_{n-1} + T_{n-2} + T_{n-3}.
• Why: every sequence starts with either a 1-step (leaving T_{n-1} ways for the rest), a 2-step (leaving T_{n-2} ways), or a 3-step (leaving T_{n-3} ways).
• Base cases (computed by hand): T₁ = 1, T₂ = 2, T₃ = 4.
• Using the recurrence: T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44, T₈ = 81, T₉ = 149.

Example: For 9 stairs, there are 149 distinct sequences of steps.

🐝 Example: Wasp colony (order 2)

• wasp_n = wasp_{n-1} + 20·wasp_{n-2}, with wasp₁ = 1, wasp₂ = 1.
• This is linear of order 2 with constants c₁ = 1, c₂ = 20.
• The excerpt notes that computing wasp(100) directly via recursion becomes very slow.

🔧 The characteristic polynomial method (order 2)

🔧 Building the characteristic polynomial

Characteristic polynomial of a_n = c₁·a_{n-1} + c₂·a_{n-2}: the degree 2 polynomial c(x) = x² - c₁·x - c₂.

• Don't confuse: the recurrence has plus signs; the characteristic polynomial has minus signs.
• The roots r₁ and r₂ are found by the quadratic formula: (c₁ ± √(c₁² + 4c₂)) / 2.

🧮 The closed-form formula

Theorem (order 2, distinct roots): If r₁ ≠ r₂, then there exist unique constants z₁ and z₂ such that

a_n = z₁·r₁ⁿ + z₂·r₂ⁿ for all n ≥ 1.

• To find z₁ and z₂, substitute n = 1 and n = 2 into the formula and solve the system:
  • a₁ = z₁·r₁ + z₂·r₂
  • a₂ = z₁·r₁² + z₂·r₂²

🌟 Example: Fibonacci numbers

• Recurrence: F_n = F_{n-1} + F_{n-2}, with F₁ = 1, F₂ = 1.
• Characteristic polynomial: c(x) = x² - x - 1.
• Roots: α = (1 + √5)/2 (the golden ratio) and α̅ = (1 - √5)/2.
• Solving for z₁ and z₂ using the initial values:
  • 1 = z₁·α + z₂·α̅
  • 1 = z₁·α² + z₂·α̅²
• Result: z₁ = 1/√5, z₂ = -1/√5.
• Closed formula: F_n = (1/√5)·αⁿ - (1/√5)·α̅ⁿ.

Example: This formula lets you compute F₁₀₀ instantly, whereas recursive computation becomes impractically slow.

🚀 Higher-order recurrences

🚀 General definition (order d)

Linear recurrence relation of order d: a_n = c₁·a_{n-1} + c₂·a_{n-2} + ... + c_d·a_{n-d}, where c_d ≠ 0.

• Requires d initial values: a₁, a₂, ..., a_d.
• The staircase problem (T_n = T_{n-1} + T_{n-2} + T_{n-3}) is order 3 with c₁ = c₂ = c₃ = 1 and initial values T₁ = 1, T₂ = 2, T₃ = 4.

🔍 Characteristic polynomial (order d)

Characteristic polynomial of order d recurrence: c(x) = x^d - c₁·x^{d-1} - c₂·x^{d-2} - ... - c_{d-1}·x - c_d.

• The polynomial has degree d, so it has d roots (call them r₁, r₂, ..., r_d).

🧮 Closed-form formula (order d)

Theorem (order d, distinct roots): If the d roots r₁, ..., r_d are all different, then there exist unique constants z₁, ..., z_d such that

a_n = z₁·r₁ⁿ + z₂·r₂ⁿ + ... + z_d·r_dⁿ for all n ≥ 1.

• To find z₁, ..., z_d, substitute n = 1, 2, ..., d into the formula and solve the resulting system of d equations.
• Don't confuse: the theorem requires distinct roots; if roots repeat, the formula changes (not covered in this excerpt).

💻 Implementing recurrences in SAGE

💻 Defining a recursive function

The excerpt shows how to define the staircase sequence T_n in SAGE:

def T(x):
    if x == 1: return 1;
    if x == 2: return 2;
    if x == 3: return 4
    else: return T(x-1) + T(x-2) + T(x-3)

• The base cases (x = 1, 2, 3) are hard-coded.
• For x ≥ 4, the function calls itself recursively.
• To print the first 10 values: [print(T(n)) for n in [1..10]].

💻 Finding roots and constants

The excerpt demonstrates the workflow for Fibonacci:

1. Define the characteristic polynomial: c = x^2 - x - 1.
2. Find roots: c.roots() returns a list; extract r₁ and r₂.
3. Set up equations for z₁ and z₂ using the initial values.
4. Solve the system: solve(eqnfibo, z1, z2).
5. Define the closed-form function and compute large values (e.g., Fibo(100)).

Why this matters: The recursive definition of T(100) or fibo(100) becomes extremely slow; the closed-form formula computes it instantly.

💻 Example: Wasp colony in SAGE

• Define the recurrence: wasp_n = wasp_{n-1} + 20·wasp_{n-2}.
• Characteristic polynomial: c(x) = x² - x - 20.
• Find roots r₁ and r₂, then solve for z₁ and z₂ using wasp₁ = 1, wasp₂ = 1.
• Check by computing wasp₃; then compute wasp₁₀₀ using the closed formula.

🧷 Key distinctions and common confusions

🧷 Recurrence vs characteristic polynomial

Aspect	Recurrence relation	Characteristic polynomial
Form	a_n = c₁·a_{n-1} + c₂·a_{n-2}	c(x) = x² - c₁·x - c₂
Signs	Addition	Subtraction
Purpose	Defines the sequence	Tool to find closed formula

Don't confuse: The characteristic polynomial is derived from the recurrence but is not the same equation.

🧷 Order and initial values

• Order d means the recurrence depends on the previous d terms.
• You need exactly d initial values to uniquely determine the sequence.
• Example: Fibonacci (order 2) needs F₁ and F₂; staircase (order 3) needs T₁, T₂, and T₃.

🧷 Distinct roots assumption

• The theorems in this excerpt assume all roots are distinct (r₁ ≠ r₂ ≠ ... ≠ r_d).
• If roots repeat, the closed-form formula changes (the excerpt mentions this is covered in Chapter 7).
• Example: If c(x) = (x - 2)², the formula involves terms like n·2ⁿ, not just 2ⁿ.

🧭 Overview

🧠 One-sentence thesis

The Tower of Hanoi game can be modeled by a nonlinear recurrence relation that yields a closed-form solution of 2^n − 1 moves to transfer n disks optimally.

📌 Key points (3–5)

• What the game is: a puzzle with 3 pegs and n disks of different sizes, where you move disks one at a time without placing a larger disk on a smaller one.
• The recurrence relation: H₁ = 1 and Hₙ = 2·Hₙ₋₁ + 1 for n ≥ 2, which is nonlinear (not solvable by the methods of section 6.2).
• Closed-form solution: Hₙ = 2^n − 1 gives the minimal number of moves for n disks.
• Common confusion: this is a nonlinear recurrence (the relation includes "+ 1" in addition to the recursive term), unlike the linear recurrences of section 6.2.
• Why it matters: the recurrence captures the optimal strategy—move all but the largest disk, move the largest, then move the rest back on top.

🎮 The game and its rules

🎮 Setup and objective

• Three pegs: left, middle, right.
• n disks: each has a hole in the center; they start stacked on the left peg in order from largest (bottom) to smallest (top).
• Goal: move all disks to another peg.

🚫 Movement constraints

• On each move, transfer exactly one disk to another peg.
• Never place a larger disk on top of a smaller one.
• Example: if you have 2 disks, you must move the small disk off the large one before you can move the large disk.

🔢 Small cases

n	Hₙ	Explanation
1	1	Just move the single disk to another peg.
2	3	Move small disk to middle peg, move big disk to right peg, move small disk on top of big disk.
3	7	(Shown in exercises.)
4	15	(Shown in exercises.)

🔁 The recurrence relation

🔁 Defining Hₙ

Hₙ: the minimal number of moves needed to finish the game for n disks.

• It is not "any number of moves that works," but the smallest number that guarantees success.
• The excerpt asks "Can the tower of Hanoi be represented as a recurrence relation?" and answers yes.

🧩 The recursive structure (Lemma 6.3.2)

The recurrence is:

• Base case: H₁ = 1
• Recursive case: Hₙ = 2·Hₙ₋₁ + 1 for n ≥ 2

Why this works (proof sketch from the excerpt):

1. By definition, it takes Hₙ₋₁ moves to transfer all the disks except the biggest one to another peg (say, the middle peg).
2. Then it takes 1 move to move the biggest disk to the empty peg (say, the right peg).
3. Then, by definition, it takes Hₙ₋₁ moves to transfer all the other disks from the middle peg on top of the biggest disk on the right peg.
4. Total: Hₙ₋₁ + 1 + Hₙ₋₁ = 2·Hₙ₋₁ + 1.

Example: for n = 2, H₂ = 2·H₁ + 1 = 2·1 + 1 = 3, which matches the manual count.

⚠️ Nonlinear vs linear

• The excerpt labels this a nonlinear recurrence.
• Don't confuse: a linear recurrence (section 6.2) has the form aₙ = c₁·aₙ₋₁ + c₂·aₙ₋₂ + … (no extra constant term that doesn't come from initial conditions).
• Here, the "+ 1" makes it nonlinear, so the methods of section 6.2 do not apply.
• Exercise 2 asks "True or False: the recursion for the Tower of Hanoi problem is a linear recurrence that can be solved with the methods of section 6.2." The answer is False.

📐 The closed-form solution

📐 Formula (Lemma 6.3.3)

Hₙ = 2^n − 1

• This gives the minimal number of moves directly without computing all previous terms.
• Example: for n = 5, H₅ = 2^5 − 1 = 32 − 1 = 31 (mentioned in exercise 6).

🔍 Proof by induction

The excerpt provides a proof:

1. Base case (n = 1): H₁ = 2^1 − 1 = 1, which is true by definition.
2. Inductive hypothesis: assume Hₙ₋₁ = 2^(n−1) − 1.
3. Inductive step: by Lemma 6.3.2, Hₙ = 2·Hₙ₋₁ + 1.
   • Substitute: Hₙ = 2·(2^(n−1) − 1) + 1 = 2^n − 2 + 1 = 2^n − 1.
4. Conclusion: the result is true by induction.

📊 Verification table

n	2^n − 1	Matches Hₙ?
1	1	Yes (H₁ = 1)
2	3	Yes (H₂ = 3)
3	7	Yes (H₃ = 7, per exercises)
4	15	Yes (H₄ = 15, per exercises)
5	31	Yes (exercise 6 mentions 31 steps)

🛠️ Exercises and strategy

🛠️ Exercises overview

The exercises ask you to:

• Compute H₅ (answer: 31).
• Verify that the recurrence is nonlinear (True/False question).
• Write out explicit move sequences for n = 3 (7 moves) and n = 4 (15 moves).
• Observe patterns: which disks does the smallest disk (d₁) rest on along the way?
• For n = 5, what rules must you follow about where d₁ and d₂ can rest to achieve 31 steps?
• Write an algorithm to solve the tower of Hanoi most efficiently.

🧠 Strategy insight

• The recurrence relation itself encodes the optimal strategy: recursively move all but the largest disk, move the largest, then recursively move the rest back.
• Don't confuse: the "+ 1" in the recurrence is the single move of the largest disk; the two Hₙ₋₁ terms are the two recursive sub-problems.
• Example: to move 3 disks optimally, you must move the top 2 disks (H₂ = 3 moves), move the largest disk (1 move), then move the top 2 disks again (H₂ = 3 moves), for a total of 3 + 1 + 3 = 7 moves.

🧭 Overview

🧠 One-sentence thesis

When drawing n non-parallel lines in the plane so that at most two lines meet at any point, the number of regions grows according to a recurrence relation that yields a closed-form quadratic formula.

📌 Key points (3–5)

• What P_n measures: the number of regions formed by drawing n non-parallel lines with at most two lines intersecting at any point.
• The recurrence relation: P_0 = 1 (no lines = one region, the whole plane) and P_n = P_(n−1) + n for n ≥ 1.
• Why the recurrence works: the nth line crosses n−1 existing lines, splitting n existing regions in two, adding n new regions.
• Closed-form formula: P_n = (n² + n + 2)/2 for all n ≥ 1, proven by induction.
• Common confusion: the nth line does not create n new regions from scratch; it splits n existing regions by crossing n−1 lines.

📐 Defining the problem

📐 Setup and constraints

• Draw lines on a piece of paper.
• Lines must be non-parallel (so every pair eventually intersects).
• At most 2 lines intersect at a point (no three lines meet at one point).
• P_n counts the resulting regions.

🔢 Small examples

The excerpt provides concrete values:

• P_1 = 2 (one line splits the plane into two regions)
• P_2 = 4 = P_1 + 2
• P_3 = 7 = P_2 + 3
• P_4 = 11 = P_3 + 4

Example: with zero lines, the entire plane is one region, so P_0 = 1.

🔁 The recurrence relation

🔁 Base case and recursive step

Recurrence relation: P_0 = 1 and P_n = P_(n−1) + n for n ≥ 1.

• Base case: no lines means one region (the whole plane), so P_0 = 1.
• Inductive step: start with n−1 lines (P_(n−1) regions). When you add the nth line, it crosses n−1 other lines (because lines are non-parallel and at most two meet at a point).

✂️ Why adding the nth line adds n regions

• The nth line intersects n−1 existing lines.
• Each intersection divides one existing region into two.
• So the nth line splits n existing regions in two, adding n new regions total.
• Hence P_n = P_(n−1) + n.

Don't confuse: the nth line does not create n brand-new regions; it splits n regions that were already there.

📊 Closed-form formula

📊 The formula

Closed form: P_n = (n² + n + 2)/2 for all n ≥ 1.

The excerpt proves this by induction.

🧮 Proof by induction

Base case (n = 1):

• P_1 = 2.
• Formula gives (1² + 1 + 2)/2 = 4/2 = 2. ✓

Inductive step:

• Assume P_n = (n² + n + 2)/2.
• Want to show P_(n+1) = ((n+1)² + (n+1) + 2)/2 = (n² + 3n + 4)/2.
• By the recurrence relation, P_(n+1) = P_n + (n+1).
• Substitute the inductive hypothesis:
  • P_(n+1) = (n² + n + 2)/2 + (n+1)
  • = (n² + n + 2)/2 + (2n + 2)/2
  • = (n² + 3n + 4)/2. ✓
• Hence the formula holds by induction.

📋 Computed values

The excerpt provides a table of values using the recurrence:

n	0	1	2	3	4	5	6	7	8	9	10
P_n	1	2	4	7	11	16	22	29	37	46	56

Example: for n = 10, the formula gives (100 + 10 + 2)/2 = 112/2 = 56.

🔄 Related problem: circles in general position

🔄 Circles instead of lines

The exercises introduce a similar problem with circles:

• Draw n circles in general position: each pair of circles intersects in exactly two points.
• Let a_n be the number of regions formed.

🔄 Key differences

• Lines: the nth line crosses n−1 lines at n−1 points, adding n regions.
• Circles: the nth circle crosses n−1 circles at 2(n−1) points, adding 2(n−1) regions.
• Recurrence for circles: a_n = a_(n−1) + 2(n−1).
• Closed form for circles: a_n = n² − n + 2 (to be shown by induction in the exercises).

Don't confuse: lines and circles have different recurrence relations because a circle intersects each other circle in two points, not one.

🧭 Overview

🧠 One-sentence thesis

Derangements count the permutations where no element remains in its original position, and they satisfy elegant recurrence relations that allow efficient computation.

📌 Key points (3–5)

• What a derangement is: a permutation where every element moves away from its initial position.
• How to count them: use recurrence relations rather than brute-force enumeration; two main formulas exist.
• Key recurrence: D_n = (n − 1)(D_{n−1} + D_{n−2}) splits the problem by whether the first element swaps with exactly one other element or participates in a larger cycle.
• Common confusion: don't confuse "permutation" (any rearrangement) with "derangement" (only rearrangements where no element stays in place).
• Why it matters: derangements model real-world scenarios where every item must change position, and they connect to the constant e through the approximation D_n ≈ n!/e.

🔢 What derangements are

🔢 Definition and basic examples

Derangement of a sequence a₁, a₂, ..., aₙ: a permutation such that no element aᵢ appears in its initial position (the i-th spot).

• Let D_n be the number of derangements of n distinct objects.
• Example: for the letters in MAT (3 letters), only 2 out of 3! = 6 permutations are derangements: ATM and TMA.
  • In ATM: M moved from position 1 to 3, A moved from 2 to 1, T moved from 3 to 2—no letter stayed.
  • Contrast with MAT itself or AMT, where at least one letter remains in its original spot.
• Example: for MATH (4 letters), 9 out of 4! = 24 permutations are derangements.

📊 Small values

n	D_n	Total permutations n!	Interpretation
1	0	1	A single element cannot move away from itself
2	1	2	Only one swap is possible
3	2	6	Two derangements out of six permutations
4	9	24	Nine derangements out of twenty-four
5	44	120	Forty-four derangements
10	1,334,961	3,628,800	Over a million derangements

• Don't confuse: D_n counts only the permutations where every element moves; the total number of permutations is n!, which includes cases where some elements stay in place.

🔄 The main recurrence relation

🔄 Formula: D_n = (n − 1)(D_{n−1} + D_{n−2})

• For n ≥ 3, the number of derangements satisfies:
  • D_n = (n − 1)(D_{n−1} + D_{n−2})
• This recurrence splits the problem into two cases based on where the first element goes.

🧩 Proof idea: where does element 1 go?

Setup: consider derangements of the sequence 1, 2, 3, ..., n.

• Let c be the position where element 1 appears.
• Since 1 cannot stay in position 1 (else not a derangement), there are n − 1 choices for c.

Case 1: elements 1 and c swap spots

• Element 1 goes to position c, and element c goes to position 1.
• Example: when n = 5 and c = 3, the derangement 34152 has 1 in position 3 and 3 in position 1.
• The remaining n − 2 elements (excluding 1 and c) must form a derangement among themselves.
• Number of ways: D_{n−2}.

Case 2: elements 1 and c do not swap

• Element 1 goes to position c, but element c does not go to position 1.
• Example: when n = 5 and c = 3, the derangement 23154 has 1 in position 3 but 3 in position 5 (not position 1).
• Now we have n − 1 elements (2, 3, ..., n) to place in n − 1 spots (positions 1, 2, ..., n, excluding position c).
• Each element is restricted from exactly one spot:
  • Element c cannot go to position 1 (to stay in Case 2).
  • Element j (j ≠ c) cannot go to position j (its original spot).
• This is exactly a derangement of n − 1 objects.
• Number of ways: D_{n−1}.

Combining the cases

• Cases 1 and 2 do not overlap.
• For a fixed choice of c, the total is D_{n−1} + D_{n−2}.
• Since there are n − 1 choices for c, the total number of derangements is (n − 1)(D_{n−1} + D_{n−2}).

📐 Alternative formulas

📐 Second recurrence: D_n = n·D_{n−1} + (−1)ⁿ

• For n ≥ 2, another recurrence relation holds:
  • D_n = n·D_{n−1} + (−1)ⁿ
• This formula is proven by induction using the first recurrence.
• It provides a simpler recursive computation when only the previous term is needed.

📐 Closed-form formula

• The number of derangements has an explicit formula:
  • D_n = n! · (1 − 1/1! + 1/2! − 1/3! + ... + (−1)ⁿ/n!)
• This is an alternating sum that resembles the Taylor series for 1/e.
• The formula can be derived using the inclusion-exclusion principle (see exercises).

🔍 Connection to e

• An interesting fact: D_n is the integer closest to n!/e.
• This means the probability that a random permutation is a derangement approaches 1/e ≈ 0.368 as n grows large.
• Example: for n = 5, D₅ = 44 and 5!/e ≈ 44.15, so the distance is about 0.15.
• Don't confuse: this is an approximation for large n; for small n, the exact values differ slightly.

🧮 Computing derangements

🧮 Using the recurrence

• Start with base cases: D₁ = 0, D₂ = 1.
• Apply the recurrence D_n = (n − 1)(D_{n−1} + D_{n−2}) iteratively:
  • D₃ = 2·(1 + 0) = 2
  • D₄ = 3·(2 + 1) = 9
  • D₅ = 4·(9 + 2) = 44
  • D₆ = 5·(44 + 9) = 265
  • And so on.

🧮 Using inclusion-exclusion

• The closed-form formula comes from counting permutations that fix at least one element, then taking the complement.
• Let W_k be the set of permutations that fix element k (i.e., element k stays in position k).
• The union of all W_k is the set of permutations that fix at least one element.
• The complement of this union is the set of derangements.
• Inclusion-exclusion gives:
  • Size of union = sum over all subsets of fixed elements, alternating signs.
  • D_n = n! − (size of union) = n! · sum from i=0 to n of (−1)ⁱ/i!.

🧮 Practical example

• For n = 4 (the word MATH):
  • Total permutations: 4! = 24.
  • Using the recurrence: D₄ = 3·(2 + 1) = 9.
  • So 9 out of 24 permutations are derangements.
  • The 9 derangements are listed: AMHT, ATHM, AHMT, TMHA, THMA, THAM, HMAT, HTMA, HTAM.

🧭 Overview

🧠 One-sentence thesis

The Catalan numbers form a celebrated combinatorial sequence defined by a simple recursion but counting an astonishing variety of structures—from lattice paths to parenthesizations—all unified by the same underlying pattern.

📌 Key points (3–5)

• Definition: The Catalan numbers C₀, C₁, C₂, … start with C₀ = 1 and satisfy the recursion C_{n+1} = C₀·C_n + C₁·C_{n-1} + … + C_n·C₀.
• Dyck paths as the key interpretation: C_n counts Dyck paths of length 2n—lattice paths from (0,0) to (n,n) using only up and right steps that stay on or above the main diagonal.
• Closed-form formula: C_n = (2n choose n) − (2n choose n+1) = 1/(n+1) · (2n choose n).
• Common confusion: Dyck paths vs arbitrary lattice paths—Dyck paths must stay on or above the diagonal at all times, not just reach (n,n).
• Remarkable universality: The same numbers count 214 different combinatorial objects, including parenthesizations, polygon triangulations, grid tilings, and binary trees.

🔢 Definition and basic values

🔢 The recursive definition

Catalan numbers: The sequence C₀, C₁, C₂, … with initial value C₀ = 1 and recurrence relation C_{n+1} = C₀·C_n + C₁·C_{n-1} + C₂·C_{n-2} + … + C_n·C₀.

• The recursion is a sum of products: each term pairs C_i with C_{n−i}.
• This structure hints at "splitting" or "decomposing" a structure of size n+1 into two smaller pieces.

📊 First ten values

Starting from n = 0, the first ten Catalan numbers are:

n	0	1	2	3	4	5	6	7	8	9
C_n	1	1	2	5	14	42	132	429	1430	4862

• The sequence grows rapidly but is determined entirely by the recursion.
• Example: C₃ = 5 means there are 5 Dyck paths of length 6, 5 ways to parenthesize 4 letters, etc.

🛤️ Dyck paths: the central interpretation

🛤️ What is a Dyck path?

Dyck path of length 2n: A path on the integer lattice grid from (0,0) to (n,n) using only up and right steps that stays on or above the main diagonal.

• "On or above the diagonal" means at every point (x,y) on the path, y ≥ x.
• The path never dips below the line y = x.
• Example: For n = 3 (length 6), there are exactly 5 such paths.

🔤 Dyck words as an alternative representation

• A Dyck path can be encoded as a Dyck word: a sequence of n U's (up steps) and n R's (right steps).
• Constraint: reading left to right, at any point the number of U's read so far must be at least the number of R's.
• Example: The five Dyck words of length 6 are UUURRR, UURURR, UURRUR, URUURR, URURUR.
• Don't confuse: any sequence of n U's and n R's is a valid word, but only those satisfying the "U's ≥ R's at all times" condition are Dyck words.

🧩 Proof that Dyck paths satisfy the Catalan recursion

The proof uses a "first return" argument:

1. Base case: There is exactly one Dyck path of length 0 (the empty path), so D₀ = 1 = C₀.
2. Recursion: Consider a Dyck path of length 2(n+1). After the first up step, the path eventually returns to the diagonal for the first time at some height i (where 1 ≤ i ≤ n+1).
3. Counting by first return height:
   • The portion from (0,0) to (i,i) is determined by a Dyck path from (0,1) to (i−1,i), giving D_{i−1} choices.
   • The remaining portion from (i,i) to (n+1,n+1) is a Dyck path of height n+1−i, giving D_{n+1−i} choices.
   • By the multiplication principle, D_{n+1,i} = D_{i−1} · D_{n+1−i}.
4. Summing over all first return heights: D_{n+1} = D₀·D_n + D₁·D_{n−1} + … + D_n·D₀, which matches the Catalan recursion.
5. Conclusion: Since D_n satisfies the same recursion and initial condition as C_n, we have D_n = C_n for all n.

🧮 Closed-form formula

🧮 The binomial formula

Theorem: The nth Catalan number is C_n = (2n choose n) − (2n choose n+1) = 1/(n+1) · (2n choose n).

• The two expressions in the formula are equivalent.
• "(2n choose n)" means the binomial coefficient: the number of ways to choose n items from 2n.
• The formula will be proven later in Section 7.6 (not included in this excerpt).
• Example: For n = 3, C₃ = 1/4 · (6 choose 3) = 1/4 · 20 = 5.

🔍 Why this formula matters

• The recursion is easy to understand but slow to compute for large n.
• The closed form allows direct calculation without computing all previous values.
• It connects Catalan numbers to binomial coefficients, a fundamental combinatorial object.

🌟 The 214 interpretations

🌟 Why Catalan numbers are celebrated

• The excerpt states there are 214 different known combinatorial descriptions of the Catalan numbers.
• All these seemingly unrelated counting problems yield the same sequence.
• This universality suggests a deep underlying structure in combinatorics.

📋 Seven interpretations when n = 3 (C₃ = 5)

The fact that C₃ = 5 is equivalent to counting:

Interpretation	Description	Example constraint
Parenthesizations	Ways to put parentheses around 4 letters	((xy)z)w, (x(yz))w, (xy)(zw), x((yz)w), x(y(zw))
Polygon triangulations	Ways to divide a convex pentagon into triangles	Non-crossing diagonals
Stair tilings	Ways to tile a stair of height 3 with rectangles	Rectangles fit the stair shape
Grid arrangements	Ways to arrange {1,…,6} in a 2×3 grid	Each row and column increasing
Non-crossing pairings	Ways to pair vertices of a hexagon	Line segments do not intersect
Binary trees	Rooted binary trees with 4 leaves	Every vertex has 0 or 2 children
Balanced sequences	Sequences of 1's and −1's of length 6	All partial sums ≥ 0, total sum = 0

🔗 Connection between interpretations

• The balanced sequences (interpretation 7) directly correspond to Dyck words: replace 1 with U and −1 with R.
• "Partial sums ≥ 0" translates to "number of U's ≥ number of R's at all times."
• Don't confuse: not all sequences with equal 1's and −1's are valid—only those where partial sums never go negative.

🎯 Related counting problems

🎯 Variations on Dyck paths

The exercises explore related problems:

• On or below the diagonal: Paths from (0,0) to (n,n) that stay on or below y = x (by symmetry, also counted by C_n).
• Strictly above the diagonal: Paths that never touch the diagonal except at endpoints (a different count).
• Ballot problem: Counting vote sequences where one candidate is never ahead until the last ballot—this is equivalent to counting certain restricted Dyck-like sequences.

🧩 Generalizing to n = 4

When n = 4, C₄ = 14. The exercises ask to enumerate all 14 instances for each interpretation:

• 14 Dyck paths of length 8
• 14 ways to parenthesize 5 letters
• 14 ways to triangulate a hexagon
• 14 ways to tile a stair of height 4
• 14 ways to arrange {1,…,8} in a 2×4 increasing grid
• 14 non-crossing pairings of an octagon
• 14 binary trees with 5 leaves
• 14 balanced sequences of length 8

Don't confuse: each interpretation counts different objects, but the count is always the same Catalan number.

🧭 Overview

🧠 One-sentence thesis

The number of ways to make change for any amount using a given set of coins can be found by multiplying polynomials where exponents encode coin values and coefficients count the number of ways to achieve each total.

📌 Key points (3–5)

• Core technique: represent each coin type as a polynomial where exponents are possible cent values, then multiply the polynomials together.
• Reading the result: the coefficient of x^N in the expanded product tells you how many ways to make N cents.
• Why exponents work: when you multiply polynomials, exponents add, which mirrors adding coin values to reach a total.
• Common confusion: the coefficient is not the amount of money—it's the number of different ways to make that amount; the exponent is the amount.
• General principle: works for any currency system and any number of each coin type, including unlimited supplies (infinite series).

🪙 The polynomial encoding technique

🪙 How to build the polynomial for one coin type

For each type of coin worth k cents, if you have m of that coin, the polynomial is 1 + x^k + x^(2k) + ··· + x^(mk).

• Each term x^(jk) represents using j coins of that type, contributing jk cents.
• The constant term 1 = x^0 represents using zero of that coin.
• Example: 6 pennies (1 cent each) → 1 + x + x² + x³ + x⁴ + x⁵ + x⁶.
• Example: 2 nickels (5 cents each) → 1 + x⁵ + x^10.

🔢 Why multiply the polynomials

• You want to combine coins from different types.
• When you multiply polynomials, exponents add: x^a · x^b = x^(a+b).
• This addition of exponents mirrors adding the cent values from different coin types.
• Example: the term x⁶ · x⁵ = x^11 represents 6 cents in pennies plus 5 cents in nickels, totaling 11 cents.

📖 Reading the expanded product

• After expanding the product, look at the coefficient of x^N.
• That coefficient is the number of distinct ways to make N cents.
• Example: in the expansion, 2x⁶ means there are 2 ways to make 6 cents (either 6 pennies + 0 nickels, or 1 penny + 1 nickel).
• If a term like x^17 is missing (coefficient 0), there are zero ways to make 17 cents with those coins.

💰 Worked examples

💰 Simple example: 6 pennies and 2 nickels

• Pennies: x⁰ + x + x² + x³ + x⁴ + x⁵ + x⁶
• Nickels: x⁰ + x⁵ + x^10
• Product expands to: x⁰ + x + x² + x³ + x⁴ + 2x⁵ + 2x⁶ + x⁷ + x⁸ + x⁹ + 2x^10 + 2x^11 + x^12 + x^13 + x^14 + x^15 + x^16
• The largest exponent is 16 (maximum you can give: 6 pennies + 2 nickels).
• The smallest exponent is 0 (give nothing).
• The coefficient of x^11 is 2, meaning two ways to give 11 cents: (6 pennies + 5 cents in nickels) or (1 penny + 10 cents in nickels).

💰 More complex example: making change for a dollar

• Coins: 6 pennies, 2 nickels, 4 dimes (10 cents each), 3 quarters (25 cents each).
• Polynomials:
  • Pennies: 1 + x + x² + x³ + x⁴ + x⁵ + x⁶
  • Nickels: 1 + x⁵ + x^10
  • Dimes: 1 + x^10 + x^20 + x^30 + x^40
  • Quarters: 1 + x^25 + x^50 + x^75
• Multiply all four and expand (using a computer algebra system like Sage).
• The coefficient of x^100 is 5, so there are 5 ways to make 100 cents (one dollar).
• The excerpt lists those 5 ways explicitly: (3 quarters + 2 dimes + 1 nickel), (3 quarters + 2 dimes + 5 pennies), etc.

💰 Non-standard currency: Harry Potter example

• Currency: 1 Knut (base unit), 1 Sickle = 29 Knuts, 1 Galleon = 493 Knuts.
• Coins: 3 Knuts, 2 Sickles, 2 Galleons.
• Polynomials:
  • Knuts: 1 + x + x² + x³
  • Sickles: 1 + x^29 + x^58
  • Galleons: 1 + x^493 + x^986
• After expansion, every coefficient is 1, meaning exactly one way to make each amount that appears (1047, 1046, 1045, 1044, 1018, ...).
• No ways to make amounts whose exponents do not appear in the expansion.

🧮 General theorem and proof

🧮 Statement of the theorem

Theorem 7.1.5: Suppose there are m types of coins of values k₁, k₂, ..., k_m, and you have n₁, n₂, ..., n_m of each type respectively. The number of ways to make change for a total value of N is the coefficient of x^N in the product (1 + x^(k₁) + x^(2k₁) + ··· + x^(n₁k₁)) · (1 + x^(k₂) + x^(2k₂) + ··· + x^(n₂k₂)) ··· (1 + x^(k_m) + x^(2k_m) + ··· + x^(n_m k_m)).

🧮 Why the theorem works (proof sketch)

• When you multiply one term from each factor, you get x^(a₁k₁) · x^(a₂k₂) ··· x^(a_m k_m) = x^(a₁k₁ + a₂k₂ + ··· + a_m k_m), where each aᵢ ≤ nᵢ.
• The exponent a₁k₁ + a₂k₂ + ··· + a_m k_m equals N exactly when using a₁ coins of type 1, a₂ coins of type 2, etc., adds up to N cents.
• Each such product corresponds to one way of making change for N.
• The number of times x^N appears in the expansion equals the number of ways to make change for N.
• Don't confuse: each individual product of terms gives one way; the sum of all those products (collected as the coefficient) counts all the ways.

🔄 Extension to unlimited coins (generating functions)

🔄 From finite to infinite

• If a cashier has an unlimited supply of each coin, the polynomials become infinite series.
• Example for unlimited pennies: 1 + x + x² + x³ + ···
• Example for unlimited nickels: 1 + x⁵ + x^10 + x^15 + ···
• These infinite series are called generating functions.

🔄 Definition of generating function

Generating function of the sequence a₀, a₁, a₂, ... is the series a₀ + a₁x + a₂x² + a₃x³ + ···, written as the sum from i=0 to infinity of aᵢxⁱ.

• The variable x is just a formal symbol; the coefficients are what matter.
• Any polynomial can be viewed as a generating function of a sequence that is eventually zero.
• Example: the sequence 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, ... has generating function 1 + x + x² + x³ + x⁴ + x⁵ + x⁶ (the polynomial for 6 pennies).

🧭 Overview

🧠 One-sentence thesis

Generating functions encode sequences as infinite series where the coefficients (not the function values) are the focus, enabling combinatorial problems like counting coin-change combinations to be solved through algebraic manipulation.

📌 Key points (3–5)

• What a generating function is: an infinite series where each coefficient represents a term in a sequence, written as a₀ + a₁x + a₂x² + a₃x³ + ···
• The variable x is formal: we don't substitute numerical values into x; it's just a "clothesline" to display the sequence of coefficients
• Polynomials are special cases: any polynomial is a generating function of a sequence that eventually becomes all zeros
• Common confusion: generating functions look like functions but aren't used as functions—we study the coefficients, not outputs from plugging in x values
• Connection to combinatorics: multiplying generating functions (like for coins) counts combinations, similar to how polynomial multiplication worked for finite supplies

🎯 What generating functions are

🎯 The basic definition

Generating function: the series sum from i=0 to infinity of aᵢxⁱ = a₀ + a₁x + a₂x² + a₃x³ + ··· for a sequence a₀, a₁, a₂, ...

• The sequence provides the coefficients
• The variable x is raised to increasing powers
• Example: the sequence 1, 2, 3, 4, 5, ... becomes 1 + 2x + 3x² + 4x³ + ···

🔢 How polynomials fit in

• Any polynomial can be viewed as a generating function of a sequence that ends in infinitely many zeros
• Example: the sequence 1, 1, 3, 0, 0, 0, ... gives 1 + x + 3x² + 0x³ + 0x⁴ + ··· = 1 + x + 3x²
• This connects finite counting problems (limited coin supplies) to infinite ones (unlimited supplies)

🪙 From coin problems to infinite series

🪙 The motivation: unlimited coins

When a cashier has an unlimited supply of each coin type, the polynomials from finite problems become infinite:

• Pennies: 1 + x + x² + x³ + ···
• Nickels: 1 + x⁵ + x¹⁰ + x¹⁵ + ···
• Dimes: 1 + x¹⁰ + x²⁰ + x³⁰ + ···
• Quarters: 1 + x²⁵ + x⁵⁰ + x⁷⁵ + ···

🔗 Recording which positions are nonzero

• The polynomial 1 + x + x² + x³ + x⁴ + x⁵ + x⁶ encodes the sequence 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, ... (six pennies available)
• The polynomial 1 + x⁵ + x¹⁰ encodes 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, ... (ones in positions 0, 5, and 10 for two nickels)
• The exponents show which amounts are possible; the coefficient shows how many ways

📚 Key examples

📚 Fibonacci sequence

• The Fibonacci sequence 1, 1, 2, 3, 5, 8, ... has generating function 1 + x + 2x² + 3x³ + 5x⁴ + 8x⁵ + ···
• The excerpt notes that by the end of the chapter, generating functions will be used to find an explicit formula for Fibonacci numbers

📚 Binomial coefficients

The sequence (n choose 0), (n choose 1), (n choose 2), ..., (n choose n) has generating function:

• (1 + x)ⁿ = (n choose 0) + (n choose 1)x + (n choose 2)x² + ··· + (n choose n)xⁿ

Combinatorial interpretation:

• n people each hold a card showing 0 or 1 (or equivalently, choose monomial 1 or x)
• The coefficient of xᵏ in (1 + x)ⁿ counts ways to choose exactly k people (those who chose x)
• This equals (n choose k)

🧵 The "clothesline" perspective

🧵 Not actually functions

• The notation A(x) = a₀ + a₁x + a₂x² + ··· is just convenient shorthand
• We do not substitute numerical values for x
• If we tried, the infinite series might not even converge for certain x values

🧵 Why the metaphor matters

Quote from Herbert Wilf: "A generating function is a clothesline on which we hang up a sequence of numbers up for display."

• The coefficients (the sequence a₀, a₁, a₂, ...) are what we're studying
• The powers of x just organize and display them
• We manipulate the series algebraically to learn about the coefficients

🧵 Variable choice is flexible

• Any variable works: the sequence 1, 2, 3, 4, 5, ... can be written as:
  • 1 + 2x + 3x² + 4x³ + ··· (in variable x)
  • 1 + 2y + 3y² + 4y³ + ··· (in variable y)
  • The variable is just a placeholder

➕ Operations on generating functions

➕ Addition of generating functions

Addition combines corresponding coefficients:

• A(x) + B(x) = (a₀ + b₀) + (a₁ + b₁)x + (a₂ + b₂)x² + ···
• Formula: sum from n=0 to infinity of (aₙ + bₙ)xⁿ

How it works:

• Line up terms with the same power of x
• Add the coefficients in each position
• Example structure: (a₀ + a₁x + a₂x² + ···) + (b₀ + b₁x + b₂x² + ···) = (a₀ + b₀) + (a₁ + b₁)x + (a₂ + b₂)x² + ···

🔄 Other operations mentioned

The excerpt notes four main operations will be covered:

1. Addition (defined above)
2. Multiplication
3. Substitution of a monomial
4. Differentiation

(The excerpt cuts off before fully explaining multiplication and the others)

🧭 Overview

🧠 One-sentence thesis

Generating functions can be manipulated through addition, multiplication, substitution of monomials, and differentiation—operations that mirror polynomial arithmetic and calculus—to produce closed-form formulas and solve counting problems.

📌 Key points (3–5)

• Four main operations: addition (add coefficients term-by-term), multiplication (use convolution of coefficients), substitution of monomials (replace x with ax^n), and differentiation (apply the power rule term-by-term).
• Geometric series identity: the infinite series 1 + x + x² + x³ + ... equals the closed form 1/(1 − x), which is the foundation for many other identities.
• Multiplication via convolution: the coefficient of x^n in A(x)B(x) is the sum a₀b_n + a₁b_(n−1) + ... + a_n b₀, called the n-th convolution.
• Common confusion: generating functions are formal power series, not actual functions—do not substitute constants or non-monomial expressions without caution; composition G(F(x)) only makes sense when F(x) has no constant term.
• Calculus rules still apply: despite being formal objects, generating functions obey the sum rule, product rule, quotient rule, and chain rule for monomials.

➕ Addition and subtraction

➕ How addition works

Addition of generating functions: add corresponding coefficients term-by-term.

• If A(x) = sum of a_n x^n and B(x) = sum of b_n x^n, then A(x) + B(x) = sum of (a_n + b_n) x^n.
• Visually: (a₀ + a₁x + a₂x² + ...) + (b₀ + b₁x + b₂x² + ...) = (a₀ + b₀) + (a₁ + b₁)x + (a₂ + b₂)x² + ...
• Example: adding 1 + x + x² + x³ + ... and 1 − x + x² − x³ + ... gives 2 + 2x² + 2x⁴ + 2x⁶ + ...

➖ Subtraction

• Subtraction works the same way: subtract corresponding coefficients.
• Any generating function minus itself equals 0 (the generating function of the sequence 0, 0, 0, ...).

✖️ Multiplication and convolution

✖️ How multiplication works

Multiplication of generating functions: multiply every term from the left factor with every term from the right factor, then collect terms with the same power of x.

• The coefficient of x^n in A(x)B(x) is a₀b_n + a₁b_(n−1) + a₂b_(n−2) + ... + a_n b₀.
• This sum is called the n-th convolution of the sequences a₀, a₁, ... and b₀, b₁, ...
• The excerpt organizes products in a table where down-and-left diagonals have the same exponent of x.

🧮 Example: squaring the sequence 1, 2, 3, 4, ...

• The generating function is 1 + 2x + 3x² + 4x³ + ...
• Squaring it: (1 + 2x + 3x² + ...)(1 + 2x + 3x² + ...)
• Collect terms along diagonals:
  • Constant term: 1
  • Coefficient of x: 2 + 2 = 4
  • Coefficient of x²: 3 + 4 + 3 = 10
  • Coefficient of x³: 4 + 6 + 6 + 4 = 20
• Result: 1 + 4x + 10x² + 20x³ + ...

🔄 Multiplying by x or x^k

• Multiplying a generating function by x shifts all coefficients one position to the right.
• If A(x) is the generating function of a₀, a₁, a₂, ..., then xA(x) has coefficients 0, a₀, a₁, a₂, ...
• In summation notation: x · (sum of a_n x^n) = sum of a_n x^(n+1) = sum (starting at m=1) of a_(m−1) x^m.
• Example: multiplying the geometric series by x gives x + x² + x³ + ... = x/(1 − x).
• Multiplying by 2x² gives 2x² + 2x³ + 2x⁴ + ... = 2x²/(1 − x).

🔑 The geometric series and closed forms

🔑 Geometric series identity

Geometric series: 1 + x + x² + x³ + ... = 1/(1 − x).

• Proof idea: multiply (1 + x + x² + ...) by (1 − x). The coefficient of x⁰ is 1, and all other coefficients are 1 − 1 = 0, so the product equals 1.
• This is the most important generating function identity.

📐 Closed form formulas

• A closed form formula is a formula that does not involve an infinite summation.
• The geometric series has closed form 1/(1 − x).
• Closed forms are useful for computing coefficients and solving problems.

🔄 Related identities

• Alternating series: 1 − x + x² − x³ + x⁴ − ... = 1/(1 + x).
• Squaring the geometric series: (1 + x + x² + ...)² = 1 + 2x + 3x² + 4x³ + ... = 1/(1 − x)².
  • This gives a closed form for the sequence 1, 2, 3, 4, ...
  • Can be derived either by multiplication (convolution) or by differentiation (see below).

🔀 Substitution of monomials

🔀 What substitution means

Monomial: a term of the form ax^n where a is a number and n is a positive integer (the degree).

• You may substitute a monomial of positive degree into a generating function identity to produce new identities.
• Example: starting with G(x) = 1 + x + x² + x³ + ... = 1/(1 − x), substitute 2x for x:
  • G(2x) = 1 + 2x + (2x)² + (2x)³ + ... = 1 + 2x + 4x² + 8x³ + ...
  • Closed form: 1/(1 − 2x).

🧩 Another example: even powers

• To find a closed form for 1 + x² + x⁴ + x⁶ + ..., substitute x² for x in the geometric series:
  • G(x²) = 1/(1 − x²).

⚠️ Warning: do not substitute constants or non-monomials

• Common confusion: do not substitute constants (like y = 1) or expressions that are not monomials (like y = 1 + x) into generating functions without caution.
• The composition G(F(x)) only makes sense when F(x) has no constant term.
• Example of what goes wrong: substituting y = 1 + x into 1 + y + y² + ... would produce an infinite series in each term, which is not well-defined.

🧮 Differentiation

🧮 How differentiation works

Differentiation of generating functions: apply the power rule term-by-term.

• Definition: d/dx (sum of a_n x^n) = sum (starting at n=1) of n·a_n x^(n−1) = sum (starting at n=0) of (n+1)·a_(n+1) x^n.
• Visually: d/dx (a₀ + a₁x + a₂x² + a₃x³ + ...) = a₁ + 2a₂x + 3a₃x² + 4a₄x³ + ...
• The constant term disappears, and each coefficient is multiplied by its original exponent.

📏 Calculus rules apply

• Even though generating functions are formal power series (not actual functions), all the usual differentiation rules from calculus still hold:
  • Sum rule: d/dx (F(x) + G(x)) = d/dx F(x) + d/dx G(x)
  • Product rule: d/dx (F(x)G(x)) = G(x)·d/dx F(x) + F(x)·d/dx G(x)
  • Quotient rule: if G(x) has a nonzero constant term, then d/dx (F(x)/G(x)) = (G(x)·d/dx F(x) − F(x)·d/dx G(x)) / G(x)²
  • Chain rule for monomials: if H(x) = d/dx F(x) and ax^n is a monomial, then d/dx F(ax^n) = ax^(n−1)·H(ax^n).

🔍 Example: deriving 1/(1 − x)²

• Start with the geometric series: 1 + x + x² + x³ + ... = 1/(1 − x).
• Differentiate both sides: 1 + 2x + 3x² + 4x³ + ... = d/dx (1/(1 − x)) = 1/(1 − x)².
• This gives another proof of the closed form for the sequence 1, 2, 3, 4, ... (previously found by multiplication).

🧪 Example: the exponential series

• Define E(x) = 1 + x + (1/2!)x² + (1/3!)x³ + ... (the generating function for the sequence 1, 1, 1/2, 1/6, ...).
• Taking the derivative: d/dx E(x) = 1 + x + (1/2!)x² + ... = E(x).
• This makes E(x) behave like the function e^x (whose derivative is itself).

📊 Summary table of operations

Operation	How it works	Example
Addition	Add corresponding coefficients	(1 + x + x² + ...) + (1 − x + x² − ...) = 2 + 2x² + 2x⁴ + ...
Multiplication	Convolution: coefficient of x^n is sum of a_k·b_(n−k) for k = 0 to n	(1 + x + x² + ...)² = 1 + 2x + 3x² + 4x³ + ...
Substitution	Replace x with a monomial ax^n	Substitute x² into 1/(1 − x) to get 1/(1 − x²)
Differentiation	Apply power rule term-by-term	d/dx (1 + x + x² + ...) = 1 + 2x + 3x² + ...

🔗 How operations connect

• Multiplication and differentiation can both be used to find the same closed form (e.g., 1/(1 − x)² for the sequence 1, 2, 3, ...).
• Substitution extends basic identities (like the geometric series) to new sequences.
• All operations preserve the algebraic structure, allowing systematic manipulation of generating functions.

🧭 Overview

🧠 One-sentence thesis

Generating functions transform recursive sequence definitions into algebraic equations that can be solved to yield explicit formulas for the sequence terms.

📌 Key points (3–5)

• Three-step method: define the generating function, solve for it using the recurrence relation, then expand to find the coefficient formula.
• Key algebraic tools: geometric series formula, partial fractions, and derivatives of power series are used to manipulate and expand generating functions.
• Common confusion: the summation index often starts at n=1 after multiplying by x, so you must adjust by subtracting and adding back the a₀ term to match the full generating function.
• Why it works: matching two different expansions of the same generating function lets you equate coefficients, revealing the explicit formula for the sequence.
• Applications: the method solves both simple linear recurrences and more complex problems like Tower of Hanoi and plane-region counting.

🔧 The three-step method

🔧 Step 1: Define the generating function

Generating function for a sequence a₀, a₁, a₂, ...: A(x) = a₀ + a₁x + a₂x² + ⋯ = sum from n=0 to infinity of aₙxⁿ.

• You assign a formal power series to the recursively defined sequence.
• This step is purely definitional—no computation yet.
• Example: for the sequence defined by a₀=3 and aₙ=2aₙ₋₁, you write A(x) = a₀ + a₁x + a₂x² + a₃x³ + ⋯.

🔧 Step 2: Solve for the generating function

• Use the recurrence relation to create an equation that A(x) satisfies.
• Key technique: multiply A(x) by x to shift indices—the coefficient of xⁿ in xA(x) is aₙ₋₁ for n≥1.
• Index adjustment: after multiplying by x, the sum starts at n=1, so subtract and add a₀ to align with the full A(x).
• Solve the resulting algebraic equation to get a closed form for A(x) in terms of x.

Example from the excerpt:

• For aₙ = 2aₙ₋₁, multiplying by 2 gives: 2xA(x) = sum from n=1 to infinity of 2aₙ₋₁xⁿ = sum from n=1 to infinity of aₙxⁿ.
• Adjust: 2xA(x) = -a₀ + a₀ + sum from n=1 to infinity of aₙxⁿ = -3 + A(x).
• Solve: (2x - 1)A(x) = -3, so A(x) = 3/(1 - 2x).

🔧 Step 3: Expand to compute coefficients

• Expand the closed form A(x) as a power series using known formulas (geometric series, derivatives, partial fractions).
• Match the expansion sum from n=0 to infinity of (formula)xⁿ with sum from n=0 to infinity of aₙxⁿ.
• Since the two series are equal, their coefficients must match: aₙ equals the formula.

Example from the excerpt:

• A(x) = 3/(1 - 2x) = 3 · sum from n=0 to infinity of (2x)ⁿ = sum from n=0 to infinity of 3·2ⁿxⁿ.
• Therefore aₙ = 3·2ⁿ for all n.

🏗️ Worked examples

🏗️ Tower of Hanoi

• Recurrence: H₁=1, Hₙ = 2Hₙ₋₁ + 1 for n≥2; set H₀=0 so the recurrence holds for all n≥1.
• Step 1: H(x) = H₀ + H₁x + H₂x² + H₃x³ + ⋯.
• Step 2: substitute the recurrence into the coefficients:
  • H(x) = H₀ + (2H₀ + 1)x + (2H₁ + 1)x² + (2H₂ + 1)x³ + ⋯.
  • Split into two sums: H(x) = 2x(H₀ + H₁x + H₂x² + ⋯) + (x + x² + x³ + ⋯).
  • Simplify: H(x) = 2xH(x) + x/(1 - x).
  • Solve: H(x)(1 - 2x) = x/(1 - x), so H(x) = x/[(1 - x)(1 - 2x)].
• Step 3: use partial fractions to write x/[(1 - x)(1 - 2x)] = a/(1 - x) + b/(1 - 2x).
  • Finding a common denominator shows a = -1 and b = 1.
  • H(x) = 1/(1 - 2x) - 1/(1 - x) = sum from n=0 to infinity of 2ⁿxⁿ - sum from n=0 to infinity of xⁿ = sum from n=0 to infinity of (2ⁿ - 1)xⁿ.
  • Therefore Hₙ = 2ⁿ - 1 for all n.

🏗️ Regions of the plane

• Recurrence: P₀=1, Pₙ = Pₙ₋₁ + n (number of regions cut by n lines, no two parallel, no three concurrent).
• Step 1: P(x) = sum from n=0 to infinity of Pₙxⁿ.
• Step 2: use the recurrence:
  • Multiply P(x) by x to get xP(x) = sum from n=1 to infinity of Pₙ₋₁xⁿ.
  • The generating function for n is sum from n=1 to infinity of nxⁿ = x/(1 - x)² (from Example 7.3.7).
  • Add: sum from n=1 to infinity of (Pₙ₋₁ + n)xⁿ = xP(x) + x/(1 - x)².
  • This equals sum from n=1 to infinity of Pₙxⁿ = P(x) - P₀.
  • Solve: P(x) - xP(x) = 1 + x/(1 - x)², so P(x)(1 - x) = 1 + x/(1 - x)², giving P(x) = 1/(1 - x) + x/(1 - x)³.
• Step 3: expand both terms:
  • 1/(1 - x) = sum from n=0 to infinity of xⁿ (geometric series).
  • For x/(1 - x)³, take the derivative of 1/(1 - x)² = sum from n=0 to infinity of (n+1)xⁿ:
    • d/dx[(1 - x)⁻²] = 2/(1 - x)³ = sum from n=1 to infinity of n(n+1)xⁿ⁻¹.
    • So 1/(1 - x)³ = (1/2)sum from n=1 to infinity of n(n+1)xⁿ⁻¹.
    • Multiply by x: x/(1 - x)³ = sum from n=1 to infinity of [n(n+1)/2]xⁿ = sum from n=0 to infinity of [n(n+1)/2]xⁿ (since the n=0 term is 0).
  • Combine: P(x) = sum from n=0 to infinity of [1 + n(n+1)/2]xⁿ = sum from n=0 to infinity of [(n² + n + 2)/2]xⁿ.
  • Therefore Pₙ = (n² + n + 2)/2 for all n.

🧰 Key algebraic tools

🧰 Geometric series formula

• Formula: 1/(1 - x) = 1 + x + x² + x³ + ⋯ = sum from n=0 to infinity of xⁿ.
• Monomial substitution: replace x with cx to get 1/(1 - cx) = sum from n=0 to infinity of cⁿxⁿ.
• Example: 1/(1 - 2x) = sum from n=0 to infinity of 2ⁿxⁿ.
• This is the most frequently used expansion in step 3.

🧰 Partial fractions

Partial fractions: a method to decompose a rational function into a sum of simpler fractions.

• Used when the closed form A(x) is a ratio of polynomials with a factored denominator.
• Example from Tower of Hanoi: x/[(1 - x)(1 - 2x)] = a/(1 - x) + b/(1 - 2x).
• Find constants a and b by adding the fractions back together (common denominator) and matching numerators.
• Each simpler fraction can then be expanded using the geometric series.

🧰 Derivatives of power series

• Technique: differentiate both sides of a known generating function identity to obtain new identities.
• Example from the excerpt:
  • Start with 1/(1 - x)² = sum from n=0 to infinity of (n+1)xⁿ.
  • Differentiate: d/dx[(1 - x)⁻²] = 2/(1 - x)³ = sum from n=1 to infinity of n(n+1)xⁿ⁻¹.
  • Rearrange: 1/(1 - x)³ = (1/2)sum from n=1 to infinity of n(n+1)xⁿ⁻¹.
  • Multiply by x: x/(1 - x)³ = sum from n=0 to infinity of [n(n+1)/2]xⁿ.
• This tool is essential for handling generating functions with higher powers in the denominator.

🔍 Common pitfalls and tips

🔍 Index alignment

• The issue: after multiplying A(x) by x, the sum starts at n=1 instead of n=0.
• The fix: subtract and add the a₀ term to rewrite the sum starting at n=0, matching the definition of A(x).
• Example: 2xA(x) = sum from n=1 to infinity of aₙxⁿ = -a₀ + a₀ + sum from n=1 to infinity of aₙxⁿ = -a₀ + A(x).
• Don't confuse: the sum from n=1 to infinity of aₙxⁿ is not equal to A(x); it equals A(x) - a₀.

🔍 Concrete handle tip

• Recommendation: compute the first few terms of the sequence using the recurrence before manipulating the generating function.
• Example: for a₀=3, aₙ=2aₙ₋₁, compute a₁=6, a₂=12, a₃=24.
• Write out: A(x) = 3 + 6x + 12x² + 24x³ + ⋯ and 2xA(x) = 6x + 12x² + 24x³ + ⋯.
• Lining up terms visually shows that 2xA(x) - A(x) = 3, making the algebra more intuitive.

🔍 Notation flexibility

• The excerpt shows two styles: using "⋯" (dot-dot-dot) notation and using summation (∑) notation.
• Both are valid; the dot-dot-dot style can be clearer for beginners because you see the pattern in the first few terms.
• The summation style is more formal and precise, especially when adjusting indices.

📊 Linear recurrences preview

📊 General linear recurrence form

• The excerpt introduces solving linear recurrences at the end (section 7.5).
• Example recurrence: a₀=0, a₁=1, aₙ = 5aₙ₋₁ - 6aₙ₋₂ for n≥2.
• The first several terms are 0, 1, 5, 19, 89, ⋯.
• Lemma result: the explicit formula is aₙ = 3ⁿ - 2ⁿ for all n.
• The proof follows the same three-step method, using the recurrence to derive a closed form for A(x), then expanding via partial fractions.

📊 Why generating functions work for recurrences

• The recurrence relation gives a rule for each coefficient in terms of previous coefficients.
• Multiplying the generating function by x (or x², etc.) shifts the indices, aligning terms so the recurrence can be applied.
• The result is an algebraic equation in A(x) that can be solved using standard algebra (factoring, partial fractions).
• Expanding the solution back into a power series reveals the explicit formula for the coefficients.

Budget: 1000000 tokens remaining. Continue with more examples or exercises if needed.

🧭 Overview

🧠 One-sentence thesis

Generating functions provide a systematic method to solve linear recurrences by converting the recurrence relation into an algebraic equation, factoring, and using partial fractions to extract an explicit formula for the sequence.

📌 Key points (3–5)

• Core technique: multiply the generating function by the reverse of the characteristic polynomial to eliminate most coefficients, leaving only initial terms.
• Partial fractions decomposition: after solving for the generating function in closed form, split it into simpler fractions that can be expanded as geometric series.
• Extracting the formula: the coefficient of x to the power n in the expanded generating function gives the explicit formula for the n-th term.
• Common confusion: the "reverse" characteristic polynomial (1 minus c₁x minus c₂x² etc.) is used, not the original characteristic polynomial (x² minus c₁x minus c₂ etc.).
• General result: when all roots of the characteristic polynomial are distinct, the solution is always a linear combination of powers of those roots.

🔧 The generating function method

🔧 Setting up the generating function

Generating function A(x) for a sequence a₀, a₁, a₂, ... is the infinite series A(x) = a₀ + a₁x + a₂x² + ... = sum from n=0 to infinity of aₙxⁿ.

• The generating function encodes the entire sequence as coefficients of a power series.
• Example: for the sequence 0, 1, 5, 19, 89, ... the generating function starts as 0 + x + 5x² + 19x³ + 89x⁴ + ...

🔄 Using the recurrence relation

• Substitute the recurrence relation into the infinite sum starting from the appropriate index.
• For aₙ = 5aₙ₋₁ - 6aₙ₋₂ (n ≥ 2), write A(x) = a₀ + a₁x + sum from n=2 to infinity of (5aₙ₋₁ - 6aₙ₋₂)xⁿ.
• Re-index the sums and factor out powers of x to express everything in terms of A(x) itself.
• This produces an algebraic equation relating A(x) to itself.

⚙️ The reverse characteristic polynomial trick

Reverse characteristic polynomial: obtained by reversing the coefficients of the characteristic polynomial; for aₙ - c₁aₙ₋₁ - c₂aₙ₋₂ - ... - cₐaₙ₋ₐ = 0, it is 1 - c₁x - c₂x² - ... - cₐxᵈ.

• The excerpt shows two equivalent approaches:
  1. Manipulate the generating function algebraically until you isolate A(x).
  2. Directly multiply A(x) by the reverse characteristic polynomial.
• Why this works: multiplying by (1 - c₁x - c₂x² - ... - cₐxᵈ) causes coefficients to cancel for n ≥ d, leaving only the initial terms.
• Example: for aₙ = 5aₙ₋₁ - 6aₙ₋₂, the characteristic polynomial is x² - 5x + 6, so the reverse is 1 - 5x + 6x².
• Don't confuse: the reverse polynomial has the constant term first (1 - ...), not the highest power first.

🧮 Solving for the explicit formula

🧮 Factoring and partial fractions

• After solving for A(x), factor the denominator into linear factors (1 - r₁x)(1 - r₂x)...(1 - rₐx), where r₁, r₂, ..., rₐ are the roots of the characteristic polynomial.
• Use partial fractions to decompose A(x) into a sum: A(x) = z₁/(1 - r₁x) + z₂/(1 - r₂x) + ... + zₐ/(1 - rₐx).
• Solve for the constants z₁, z₂, ..., zₐ by combining fractions and comparing coefficients.

Example from the excerpt:

• For A(x) = x/((1 - 2x)(1 - 3x)), write A(x) = r/(1 - 2x) + s/(1 - 3x).
• Combine to get (r + s - (3r + 2s)x)/((1 - 2x)(1 - 3x)) = x/((1 - 2x)(1 - 3x)).
• Equate numerators: r + s = 0 and 3r + 2s = -1, giving r = -1 and s = 1.

📐 Expanding as geometric series

• Each fraction 1/(1 - rx) expands as the geometric series 1 + rx + r²x² + r³x³ + ... = sum from n=0 to infinity of rⁿxⁿ.
• Multiply by the constant from partial fractions: z/(1 - rx) = sum from n=0 to infinity of z·rⁿxⁿ.
• Add all the expanded series together.
• The coefficient of xⁿ in the final sum is the explicit formula for aₙ.

Example: A(x) = -1/(1 - 2x) + 1/(1 - 3x) expands to sum from n=0 to infinity of (3ⁿ - 2ⁿ)xⁿ, so aₙ = 3ⁿ - 2ⁿ.

🌟 Worked examples

🌟 Example: aₙ = 5aₙ₋₁ - 6aₙ₋₂

• Initial conditions: a₀ = 0, a₁ = 1.
• Sequence: 0, 1, 5, 19, 89, ...
• After manipulation, A(x) = x/(1 - 5x + 6x²) = x/((1 - 2x)(1 - 3x)).
• Partial fractions: A(x) = -1/(1 - 2x) + 1/(1 - 3x).
• Explicit formula: aₙ = 3ⁿ - 2ⁿ for all n.

🌟 Example: Fibonacci numbers

• Initial conditions: F₀ = 0, F₁ = 1, recurrence Fₙ = Fₙ₋₁ + Fₙ₋₂ for n ≥ 2.
• Generating function: G(x) = sum from n=0 to infinity of Fₙxⁿ.
• Multiply by reverse characteristic polynomial (1 - x - x²): G(x)(1 - x - x²) = x.
• Solve: G(x) = x/(1 - x - x²).
• Factor denominator: (1 - ax)(1 - bx) where a = (1 + √5)/2 and b = (1 - √5)/2.
• Partial fractions: G(x) = (1/√5)/(1 - ax) - (1/√5)/(1 - bx).
• Explicit formula: Fₙ = (1/√5)·((1 + √5)/2)ⁿ - (1/√5)·((1 - √5)/2)ⁿ.

📜 General theorem for distinct roots

📜 Statement of the theorem

Theorem 7.5.5: If r₁, ..., rₐ are the distinct roots of the characteristic polynomial of the recurrence aₙ = c₁aₙ₋₁ + c₂aₙ₋₂ + ... + cₐaₙ₋ₐ, then there exist unique constants z₁, ..., zₐ such that aₙ = z₁r₁ⁿ + z₂r₂ⁿ + ... + zₐrₐⁿ for all n ≥ 1.

• The theorem applies only when all roots are different (distinct).
• The constants z₁, ..., zₐ are determined uniquely by the initial conditions.

📜 Proof outline

1. Define the reverse polynomial r(x) = 1 - c₁x - c₂x² - ... - cₐxᵈ.
2. Multiply the generating function A(x) by r(x); all coefficients from xᵈ onward cancel, leaving r(x)A(x) = p(x) for some polynomial p(x).
3. Show that r(x) factors as (1 - r₁x)(1 - r₂x)...(1 - rₐx) by relating it to the characteristic polynomial c(x) = (x - r₁)(x - r₂)...(x - rₐ).
4. Key identity: r(x) = xᵈ·c(1/x), which after simplification gives the factorization.
5. Use partial fractions to decompose A(x) = p(x)/r(x) into z₁/(1 - r₁x) + ... + zₐ/(1 - rₐx).
6. Expand each term as a geometric series; the coefficient of xⁿ is z₁r₁ⁿ + ... + zₐrₐⁿ.

📜 Why the reverse polynomial works

• The characteristic polynomial c(x) = xᵈ - c₁xᵈ⁻¹ - ... - cₐ has roots r₁, ..., rₐ.
• The reverse polynomial r(x) = 1 - c₁x - c₂x² - ... - cₐxᵈ is related by r(x) = xᵈ·c(1/x).
• Substituting c(x) = (x - r₁)...(x - rₐ) into this identity and simplifying yields r(x) = (1 - r₁x)...(1 - rₐx).
• This factorization is essential for partial fractions decomposition.

🧭 Overview

🧠 One-sentence thesis

The Catalan numbers can be derived using generating functions by recognizing that their recursive definition corresponds to squaring the generating function, leading to an explicit closed formula.

📌 Key points (3–5)

• Recursive definition: Each Catalan number is defined by a sum involving products of earlier Catalan numbers: C_{n+1} = C_0·C_n + C_1·C_{n-1} + ... + C_n·C_0, with C_0 = 1.
• Generating function approach: The recursion translates into the equation C(x)² = (C(x) - C_0)/x, which becomes a quadratic equation solvable by the quadratic formula.
• Generalized binomial coefficients: To expand the square root in the solution, the text extends binomial coefficients to rational exponents, not just positive integers.
• Common confusion: When solving C(x) = (1 ± √(1-4x))/(2x), you must choose the negative sign so the constant term cancels and all terms are divisible by 2x.
• Explicit formula: The final result is C_n = (1/(n+1)) · (2n choose n), derived by expanding the generating function and extracting coefficients.

🔢 The Catalan recursion and generating function setup

🔢 Recursive definition

The Catalan numbers are defined by the recursion C_{n+1} = C_0·C_n + C_1·C_{n-1} + C_2·C_{n-2} + ... + C_n·C_0, with initial condition C_0 = 1.

• This means each new Catalan number is built from products of all earlier pairs.
• Example: C_1 = C_0·C_0, C_2 = C_0·C_1 + C_1·C_0, C_3 = C_0·C_2 + C_1·C_1 + C_2·C_0.

📊 Defining the generating function

The generating function is:

• C(x) = C_0 + C_1·x + C_2·x² + C_3·x³ + ...
• This packages all Catalan numbers into a single formal power series.

🧮 Solving for the generating function

🧮 Recognizing the product structure

When you multiply C(x) by itself:

• C(x)² = C(x) · C(x) = C_0·C_0 + (C_0·C_1 + C_1·C_0)·x + (C_0·C_2 + C_1·C_1 + C_2·C_0)·x² + ...
• The coefficient of x^n in C(x)² exactly matches the right-hand side of the recursion for C_{n+1}.
• Therefore: C(x)² = C_1 + C_2·x + C_3·x² + C_4·x³ + ...

🔧 Manipulating to isolate C(x)

To align subscripts with powers of x:

• Multiply both sides by x: x·C(x)² = C_1·x + C_2·x² + C_3·x³ + ...
• The right side is almost C(x), but missing the C_0 term.
• Add C_0 to both sides: x·C(x)² + C_0 = C_0 + C_1·x + C_2·x² + ... = C(x).
• Since C_0 = 1, this becomes: x·C(x)² + 1 = C(x), or x·C(x)² - C(x) + 1 = 0.

🎯 Applying the quadratic formula

Treating this as a quadratic in C(x):

• C(x) = (1 ± √(1 - 4x)) / (2x)
• Which sign to choose? Use the negative sign (minus) so that the constant term 1 in the numerator cancels and all terms become divisible by 2x.
• Don't confuse: both signs solve the quadratic algebraically, but only the minus sign gives a valid power series starting with C_0 = 1.

🔬 Expanding the square root

🔬 Interpreting √(1 - 4x) as a generating function

To find coefficients, think of √(1 - 4x) as "the generating function whose square is 1 - 4x."

• Write (a_0 + a_1·x + a_2·x² + ...)² = 1 - 4x.
• Expand the left side: a_0² + (a_0·a_1 + a_1·a_0)·x + (a_0·a_2 + a_1·a_1 + a_2·a_0)·x² + ... = 1 - 4x.
• Match coefficients term by term:
  • From x⁰: a_0² = 1, so a_0 = 1 (take the positive root by convention).
  • From x¹: 2·a_0·a_1 = -4, so a_1 = -2.
  • From x²: 2·a_0·a_2 + a_1² = 0, so a_2 = -2.
• Each coefficient a_n is uniquely determined by earlier coefficients.

📐 Generalized binomial coefficients

Definition: For any rational number α and integer k, define (α choose k) = α(α-1)(α-2)···(α-k+1) / k! if k ≠ 0, and (α choose 0) = 1.

• When α is a positive integer n, this matches the ordinary binomial coefficient (n choose k).
• Example: (1/2 choose 3) = (1/2)(-1/2)(-3/2) / 3! = (3/8) / 6 = 1/16.

🧪 Generalized binomial theorem

Theorem: As a generating function, (1 + x)^α = Σ_{k=0}^∞ (α choose k)·x^k.

• This extends the ordinary binomial theorem to rational exponents.
• Example: Set α = 1/2 and substitute -4x for x:
  • √(1 - 4x) = Σ_{k=0}^∞ (1/2 choose k)·(-4x)^k = Σ_{k=0}^∞ (1/2 choose k)·(-1)^k·4^k·x^k.

🎓 Deriving the explicit formula

🎓 Extracting the coefficient of x^n

Starting from C(x) = (1 - √(1 - 4x)) / (2x):

• Substitute the expansion: C(x) = (1 - (Σ_{k=0}^∞ (1/2 choose k)·(-1)^k·4^k·x^k)) / (2x).
• Separate the k=0 term: C(x) = (1 - (1 + Σ_{k=1}^∞ (1/2 choose k)·(-1)^k·4^k·x^k)) / (2x).
• Simplify: C(x) = (-Σ_{k=1}^∞ (1/2 choose k)·(-1)^k·4^k·x^k) / (2x).
• Substitute n = k - 1: C(x) = Σ_{n=0}^∞ (1/2 choose n+1)·(-1)^{n+1}·4^{n+1}·x^{n+1} / (2x).
• Cancel x and simplify: C(x) = Σ_{n=0}^∞ (1/2 choose n+1)·(-1)^n·2^{2n+1}·x^n / 2.

🏆 Final closed formula

From Example 7.6.3, (1/2 choose n+1) = (2n)!·(-1)^n / (2^{2n+1}·n!·(n+1)!).

• Substitute into the coefficient of x^n:
  • C_n = [(2n)!·(-1)^n / (2^{2n+1}·n!·(n+1)!)] · [(-1)^n·2^{2n+1}] / 2.
  • The (-1)^n and 2^{2n+1} terms cancel.
  • C_n = (2n)! / (n!·(n+1)!) = (1/(n+1)) · (2n)! / (n!·n!) = (1/(n+1)) · (2n choose n).

Corollary: The n-th Catalan number is C_n = (1/(n+1)) · (2n choose n).

This completes the derivation using generating functions.

🧭 Overview

🧠 One-sentence thesis

This chapter provides practice problems on generating functions, including coin-change polynomials, closed-form coefficients, recursive sequences, and an introduction to partition theory using infinite product formulas.

📌 Key points (3–5)

• Coin-change problems: model the number of ways to make change by multiplying polynomials whose exponents represent coin values.
• Coefficient extraction: find coefficients in generating functions with closed forms like 1/(1 − 3x) or x²/(1 − 3x).
• Recursive sequences: use generating functions to find closed formulas for sequences defined by recurrence relations (e.g., Fibonacci-like sequences).
• Partitions: a partition of n is a way to write n as a sum of positive integers where order doesn't matter; p(n) counts all partitions, q(n) counts partitions with distinct parts.
• Common confusion: in partitions, (2, 2, 1) and (2, 1, 2) are the same partition because order doesn't matter; parts are written in weakly decreasing order by convention.

💰 Coin-change and polynomial methods

💰 Modeling coin combinations

• Problem setup: you have 8 pennies, 3 nickels, and 1 dime; you want to give a subset to a friend.
• For each coin type, write a polynomial whose exponents are the possible values you could give:
  • Pennies: exponents 0, 1, 2, …, 8 (since you have 8 pennies, each worth 1 cent).
  • Nickels: exponents 0, 5, 10, 15 (since you have 3 nickels, each worth 5 cents).
  • Dimes: exponents 0, 10 (since you have 1 dime worth 10 cents).
• Example polynomial for dimes: f_d = 1 + x^10.

🧮 Multiplying polynomials to count ways

• Multiply the three polynomials f_p, f_n, f_d to get a product F.
• The coefficient of x^k in F tells you how many ways you can give k cents.
• Example questions:
  • What is the degree of F? (The largest power of x that appears.)
  • How many ways can you give 17 cents? (Look at the coefficient of x^17.)
  • How many ways can you give 20 cents? (Look at the coefficient of x^20.)
• If the number of dimes changes to 2, the polynomial for dimes becomes f_d = 1 + x^10 + x^20, which changes the degree and the coefficients.

🖥️ Using Sage for computation

• The excerpt suggests using Sage (sagecell.sagemath.org) to define and expand the polynomials.
• Useful commands:
  • Define a polynomial ring: R = PolynomialRing(RR, x).
  • Expand the product: F = expand(fp * fn * fd).
  • Extract a specific coefficient: F.coefficients()[17] returns the coefficient of x^17 and the number 17.
• Don't confuse: ending a Sage cell with F returns the entire polynomial; using F.coefficients() returns only the coefficients.

🔢 Coefficient extraction from closed forms

🔢 Geometric series and powers

• Problem 3: What is the coefficient of x^4 in 1/(1 − 3x)?
  • Use the geometric series formula: 1/(1 − 3x) = sum from k=0 to infinity of (3x)^k = sum of 3^k x^k.
  • The coefficient of x^4 is 3^4 = 81.
• Problem 4: What is the coefficient of x^4 in x²/(1 − 3x)?
  • Rewrite: x²/(1 − 3x) = x² · (sum of 3^k x^k) = sum of 3^k x^(k+2).
  • For x^4, we need k+2 = 4, so k = 2; the coefficient is 3^2 = 9.

🔢 Squared denominators

• Problem 5: What is the coefficient of x^4 in 1/(1 − 3x)²?
  • Use the formula for (1 − r)^(−2): the coefficient of x^n is (n+1) r^n.
  • Here r = 3, so the coefficient of x^4 is (4+1) · 3^4 = 5 · 81 = 405.

🔢 Fibonacci generating functions

• Problem 6: Let F(x) = F_0 + F_1 x + F_2 x² + … be the generating function of the Fibonacci numbers (F_0 = 0, F_1 = 1, F_n = F_(n−1) + F_(n−2) for n ≥ 2).
  • Compute the coefficient of x^5 in 1/(1 − x F(x)).
  • This requires expanding 1/(1 − x F(x)) as a geometric series in x F(x) and extracting the x^5 term.
• Problem 7: What is the coefficient of x^5 in d/dx F(x)?
  • Differentiate F(x) term by term: d/dx (F_n x^n) = n F_n x^(n−1).
  • The coefficient of x^5 in d/dx F(x) is 6 F_6 (since the x^6 term in F(x) becomes 6 F_6 x^5 after differentiation).

🔁 Recursive sequences and closed formulas

🔁 Finding closed forms for recurrences

• Problem 8: Find a closed formula for the generating function of the sequence defined by b_0 = 5, b_1 = 12, and b_n = 5 b_(n−1) + 6 b_(n−2) for all n ≥ 2.
  • Let B(x) = sum of b_n x^n.
  • Multiply the recurrence by x^n and sum to relate B(x) to itself.
  • Solve for B(x) to get a closed form (typically a rational function).

🔁 Cumulative sequences

• Problem 9: Find a closed formula for the generating function of the sequence defined by e_0 = 1 and e_n = e_(n−1) + e_(n−2) + … + e_0 for all n ≥ 1.
  • Notice that e_n is the sum of all previous terms.
  • Let E(x) = sum of e_n x^n.
  • The recurrence e_n = sum from k=0 to n−1 of e_k implies E(x) = 1 + x E(x) / (1 − x) (or a similar relation).
  • Solve for E(x).

🔁 Proving identities with generating functions

• Problem 10: Prove, using generating functions, that sum from k=0 to n of k · (n choose k) = n · 2^(n−1).
  • Consider the generating function (1 + x)^n = sum of (n choose k) x^k.
  • Differentiate both sides with respect to x: n (1 + x)^(n−1) = sum of k (n choose k) x^(k−1).
  • Set x = 1 to get n · 2^(n−1) = sum of k (n choose k).

🔁 Fibonacci expansion

• Problem 11: The generating function for the Fibonacci numbers is sum of F_n x^n = x / (1 − x − x²).
  • Rewrite as x / (1 − (x + x²)) and use the geometric series formula: 1 / (1 − r) = sum of r^k.
  • Here r = x + x², so x / (1 − (x + x²)) = x · sum of (x + x²)^k.
  • Expand (x + x²)^k and collect terms to prove that F_n = sum from k=0 to n−1 of (n − k − 1 choose k).

🧩 Partitions and their generating functions

🧩 What is a partition?

Partition of a positive integer n: a way of writing n as a sum of other positive integers, where the order of the summands doesn't matter.

• The summands are called the parts of the partition.
• Convention: write parts in weakly decreasing order: λ_1 ≥ λ_2 ≥ … ≥ λ_k.
• Example: since 5 = 2 + 1 + 2, the tuple (2, 2, 1) is a partition of 5 (written in decreasing order).

🧩 Counting partitions

• p(n): the number of partitions of n (all partitions).
• q(n): the number of partitions of n whose parts are all distinct.
• Example: the seven partitions of 5 are:
  • (5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1).
  • So p(5) = 7.
  • The partitions with distinct parts are (5), (4, 1), (3, 2), so q(5) = 3.
• Example: p(6) = 11 and q(6) = 4.

🧩 Generating function for partitions

• The generating function for the partition numbers p(n) has an infinite product formula:
  • sum from n=0 to infinity of p(n) x^n = 1 / ((1 − x) · (1 − x²) · (1 − x³) · …).
• Each factor 1/(1 − x^k) corresponds to the choice of how many parts of size k to include in a partition.
• Don't confuse: this is an infinite product, not a finite polynomial product like the coin-change problems.

🧩 Distinct parts

• For partitions with distinct parts (counted by q(n)), the generating function is different (the excerpt is cut off, but typically involves products of (1 + x^k) instead of 1/(1 − x^k)).

🧭 Overview

🧠 One-sentence thesis

Partitions of integers can be enumerated using generating functions expressed as infinite products, and their Young diagrams reveal deep combinatorial identities and bijections.

📌 Key points (3–5)

• What a partition is: a way to write a positive integer as a sum of positive integers where order doesn't matter, with parts written in weakly decreasing order.
• Generating functions for partitions: the number of partitions p(n) has generating function equal to an infinite product of terms 1/(1 − x^i), while distinct-part partitions q(n) correspond to the product of (1 + x^i).
• Convergence of infinite products: an infinite product of generating functions converges when the coefficients of partial products stabilize; not all infinite products converge.
• Common confusion: partitions vs compositions—partitions ignore order (e.g., (2,2,1) and (2,1,2) are the same partition), while the parts must be arranged in weakly decreasing order as a canonical form.
• Young diagrams and bijections: visual representations of partitions as stacks of boxes reveal connections to lattice paths, Catalan numbers, and self-conjugate partitions.

🔢 Partition fundamentals

🔢 Definition and notation

A partition of a positive integer n is a way of writing n as a sum of other positive integers, where the order of the summands doesn't matter.

• The summands are called parts of the partition.
• Notation: λ = (λ₁, λ₂, ..., λₖ) represents the partition n = λ₁ + λ₂ + ⋯ + λₖ.
• Parts are ordered in weakly decreasing order: λ₁ ≥ λ₂ ≥ ⋯ ≥ λₖ.
• Example: Since 5 = 2 + 1 + 2, the tuple (2, 2, 1) is a partition of 5 (parts reordered as 2 ≥ 2 ≥ 1).

📊 Counting partitions

• p(n): the number of partitions of n (all partitions).
• q(n): the number of partitions of n whose parts are all distinct.
• Example: The seven partitions of 5 are (5), (4,1), (3,2), (3,1,1), (2,2,1), (2,1,1,1), (1,1,1,1,1), so p(5) = 7.
• Among these, only (5), (4,1), and (3,2) have all distinct parts, so q(5) = 3.

🔄 Order matters for canonical form

• Don't confuse: The partition (2,2,1) is the same as (2,1,2) or (1,2,2) because order doesn't matter in the definition.
• However, we write parts in weakly decreasing order as a canonical choice to avoid counting the same partition multiple times.

∞ Infinite products and generating functions

∞ Generating function for p(n)

The generating function for partition numbers has an infinite product formula:

Sum from n=0 to ∞ of p(n)x^n = 1/(1−x) · 1/(1−x²) · 1/(1−x³) · 1/(1−x⁴) · ⋯

• Each factor 1/(1−x^i) corresponds to allowing parts of size i in the partition.
• The video explains this informally; the excerpt explores rigorous details.

🔍 Convergence of infinite products

An infinite product of generating functions F₁(x)F₂(x)F₃(x)⋯ converges to G(x) = Σbᵢx^i if the partial products Pₘ(x) = F₁(x)⋯Fₘ(x) have coefficients that eventually match those of G(x).

• "Eventually match" means: for all i, there exists N such that if m > N, the coefficient of x^i in Pₘ(x) equals bᵢ.
• If the product converges to G(x), we say the infinite product equals G(x).

⚠️ When products don't converge

Example from Question 7.8.7: If all Fᵢ(x) = 1 + x, then:

• P₁(x) = 1 + x
• P₂(x) = (1 + x)²
• P₃(x) = (1 + x)³

The coefficient of x in Pₘ(x) is m (from the binomial expansion), which grows without bound as m increases. This pattern does not stabilize, so the infinite product (1+x)(1+x)(1+x)⋯ does not converge to any generating function.

✅ When products do converge

Example from Question 7.8.8: Define Fᵢ(x) = 1 + x^(2^(i−1)), so:

• F₁(x) = 1 + x
• F₂(x) = 1 + x²
• F₃(x) = 1 + x⁴
• F₄(x) = 1 + x⁸

The partial products expand to include all possible sums of distinct powers of 2. The infinite product (1+x)(1+x²)(1+x⁴)(1+x⁸)⋯ converges because each coefficient of x^i stabilizes once enough factors are included to represent i in binary.

🎯 Special partition generating functions

🎯 Distinct-part partitions

From Question 7.8.9:

Product from i=1 to ∞ of (1 + x^i) = Sum from n=0 to ∞ of q(n)x^n

• Each factor (1 + x^i) represents the choice: either include part i once (contributing x^i) or don't include it (contributing 1).
• This counts partitions where all parts are distinct.

🎯 Even-part partitions

From Question 7.8.10:

Product from i=1 to ∞ of 1/(1−x^(2i)) = Sum of E(n)x^n

where E(n) is the number of partitions of n that use only even parts.

• Each factor 1/(1−x^(2i)) allows using the even part 2i any number of times (0, 1, 2, ...).

🔗 A remarkable identity

Question 7.8.11 asks to multiply the two products above:

[Product of (1 + x^i)] · [Product of 1/(1−x^(2i))]

By factoring 1 − x^(2i) = (1 − x^i)(1 + x^i), the (1 + x^i) terms cancel, leaving:

Product from i=1 to ∞ of 1/(1−x^i)

This is the generating function for all partitions p(n). Comparing coefficients yields a combinatorial identity relating q(n), E(n), and p(n).

📐 Young diagrams and visual combinatorics

📐 Definition and construction

The Young diagram of a partition λ is the left-justified stack of boxes in which the i-th row from the bottom has λᵢ boxes.

• Visual representation: each part corresponds to a row of boxes.
• Example: The partition (3,2) of 5 has a Young diagram with 3 boxes in the bottom row and 2 boxes in the second row.
• The seven partitions of 5 correspond to seven distinct Young diagrams.

🔄 Transpose (conjugate) partitions

The transpose or conjugate of a partition is the partition formed by reflecting the Young diagram about the diagonal line going southwest-northeast.

• Reflection swaps rows and columns.
• Example: Two partitions shown in the excerpt are conjugates of each other (specific diagrams not reproduced here).
• Question 7.8.18 asks which partitions of 5 are self-conjugate (equal to their own transpose).

🔗 Connection to lattice paths

Question 7.8.14: Counting Young diagrams that fit inside a 3×4 grid is equivalent to counting lattice paths from the top-left corner to the bottom-right corner.

• Each Young diagram corresponds to a path that stays on or below the diagram's boundary.
• This connects partition enumeration to earlier material on lattice path counting.

🌟 Connection to Catalan numbers

Question 7.8.15: The number of partitions (λ₁, ..., λₙ) with n − i < λᵢ ≤ n for all i equals the Catalan number Cₙ.

• Hint: The Young diagram's border corresponds to a Dyck path (a lattice path that never goes below the diagonal).
• This reveals a deep connection between partitions and Catalan structures.

🔄 Self-conjugate partitions and odd parts

Question 7.8.19: There is a bijection between:

• Self-conjugate partitions of n (partitions equal to their own transpose).
• Partitions of n having all distinct odd parts.

This bijection is a classic result in partition theory, showing that two seemingly different partition classes are equinumerous.

🧭 Overview

🧠 One-sentence thesis

Graphs provide a simplified data structure that represents relationships between objects through vertices and edges, enabling powerful tools to solve optimization and pattern problems across mathematics and computer science.

📌 Key points (3–5)

• What a graph is: a collection of vertices (dots) and edges (line segments) that describe adjacency (connections or relationships) between vertices.
• Why graphs matter: they simplify complex maps and networks, making it easier to solve optimization problems and reveal patterns.
• Standard restrictions: unless stated otherwise, graphs in this book do not allow self-loops (edges from a vertex to itself) or multiple edges between the same pair of vertices.
• Common confusion: the same graph can be drawn in different ways—a graph is defined only by its vertex set and edge set, not by how it looks on paper.
• Key examples: complete graphs (every pair connected), path graphs (vertices in a line), and cycle graphs (vertices forming a loop).

🎯 What graphs represent

🎯 Real-world simplification

• The excerpt motivates graphs with a bus route problem: students want to travel between campus locations (Corbett Hall, Library, Weber, Recreation Center) during a hailstorm.
• A map with all details is confusing; a graph strips away unnecessary information.
• Vertices represent locations (bus stops).
• Edges represent direct routes between stops.
• Example: Corbett Hall and Moby Arena are adjacent bus stops, so they are connected by an edge.

🌐 Social networks

• The excerpt also describes a social network graph.
• Each vertex corresponds to a person (or at least a mammal, the excerpt jokes).
• An edge between two people means they have met before.
• Application question: "Which person would you want to contact first?" to collect votes—this hints at centrality and influence in the network.

📐 Formal definitions

📐 Graph, vertices, edges

Graph G = (V, E): consists of a finite set V of vertices (or nodes) and a finite set E of edges, where each edge e ∈ E is of the form e = {u, v} with u, v ∈ V.

• An edge {u, v} is often written as uv.
• Unless mentioned otherwise, uv is the same as vu (order does not matter).
• The graph is defined only by its vertex set and edge set, not by how it is drawn.

🚫 Standard restrictions

The excerpt emphasizes two default rules (Remark 8.1.2):

1. No self-loops: an edge uu (starting and ending at the same vertex u) is not allowed.
2. No multiple edges: at most one edge between any pair of vertices {u, v}.

• These restrictions are automatic because each edge e is a set (so {u, u} is not valid) and E is a set (so duplicate edges are not allowed).
• Later chapters will consider more general graphs that do allow self-loops and multiple edges.

🔗 Adjacency

Adjacent vertices u and v: two vertices are adjacent if there is an edge uv connecting them. If u and v are adjacent, we write u ∼ v.

• Adjacency describes a direct connection or relationship.
• Example: in the bus route graph, Corbett Hall ∼ Moby Arena because there is a direct bus route.

🏷️ Labeling

• A labeling assigns numbers to each vertex.
• If a graph has n vertices, labels are always 0 through n − 1, each used exactly once.
• The same graph can be drawn in different ways; what matters is the vertex and edge sets, not the visual layout.

🏗️ Common graph families

🏗️ Complete graphs K_n

Complete graph K_n: the graph with n vertices having an edge between every pair of distinct vertices.

• Every possible pair of vertices is connected.
• The excerpt gives examples: K₁, K₂, K₃, K₄.
• How many edges? K_n has "n choose 2" edges, because there are that many pairs of distinct vertices.
• Example (from the excerpt): What is the largest number of edges a graph with 9 vertices can have? Answer: "9 choose 2" edges, which is the complete graph K₉.

🛤️ Path graphs P_n

Path graph P_n: the graph with n vertices and n − 1 edges that can be drawn so that all vertices and edges lie on a straight line.

• Vertices are arranged in a line, each connected to the next.
• Example: P₅ has five vertices and four edges.
• The bus route in Figure 8.2 contains a path graph P₆ connecting Moby Arena → Corbett Hall → Library → CSU Transit Center → Engineering → MAX Station.

🔄 Cycle graphs C_n

Cycle graph C_n (for n ≥ 3): the graph with n edges and n vertices such that every vertex is adjacent to exactly two others.

• Vertices form a closed loop.
• Each vertex has exactly two neighbors.
• The excerpt defines cycle graphs but does not provide a worked example in the given text.

🧮 Counting edges and graphs

🧮 Maximum edges

• Question: What is the largest number of edges a graph with 5 vertices can have?
• Answer: the complete graph K₅ has "5 choose 2" edges.
• General rule: a graph with n vertices can have at most "n choose 2" edges (no self-loops, no multiple edges).

🧮 Counting distinct graphs

The excerpt includes exercises that ask:

1. Labeled vertices: How many different graphs (V, E) are there if the vertices are labeled (e.g., V = {v₁, v₂, v₃, v₄})? In this case, the graph with edge v₁v₂ is different from the graph with edge v₂v₃.
2. Unlabeled vertices: How many different graphs are there if vertices are not labeled? In this case, all graphs with 4 vertices and one edge are considered the same.

• Don't confuse: labeled vs. unlabeled graphs count differently because labeling distinguishes otherwise identical structures.

🧮 Graphs with a fixed number of edges

• Exercise: How many different graphs (V, E) with exactly three edges are there if V has 5 vertices?
  • (a) Labeled vertices.
  • (b) Unlabeled vertices.
• The excerpt does not provide answers, but the distinction between labeled and unlabeled is key.

🔍 Drawing and recognizing graphs

🔍 Same graph, different drawings

• The excerpt emphasizes: "A graph is defined only by its set of vertices and its set of edges; the same graph may be drawn in different ways."
• Two graphs below (in the excerpt) are the same graph, just drawn differently in the plane.
• Don't confuse: visual appearance vs. structural identity. If two drawings have the same vertex set and edge set, they are the same graph.

🔍 Ignoring edge crossings (for now)

• In the social network example (Figure 8.3), "The one intersection of edges that does not have a dot means nothing and can be ignored."
• Edge crossings in a drawing do not create new vertices unless explicitly marked.
• Later (Chapter 11), the excerpt mentions graphs that can be drawn with no crossing edges, but for now crossings are just artifacts of the drawing.

🧭 Overview

🧠 One-sentence thesis

Certain graph structures appear so frequently in applications that they have standard names—complete graphs, paths, cycles, and bipartite graphs—each capturing a different pattern of connections.

📌 Key points (3–5)

• Complete graphs (Kₙ): every pair of vertices is connected; the maximum number of edges for n vertices is "n choose 2."
• Path and cycle graphs (Pₙ, Cₙ): paths are vertices arranged in a line; cycles close the path into a loop where every vertex has exactly two neighbors.
• Bipartite graphs: vertices split into two groups (left and right) with edges only between groups, never within a group—useful for matching problems.
• Common confusion: a cycle graph Cₙ requires n ≥ 3 and every vertex must have degree exactly 2; don't confuse it with a path, which has endpoints of degree 1.
• Why it matters: recognizing these named graphs helps model real-world networks like bus routes, ordering patterns, and matching scenarios.

🔗 Complete graphs

🔗 What a complete graph is

Complete graph Kₙ: the graph with n vertices having an edge between every pair of distinct vertices.

• Every possible connection exists.
• Example: K₄ has 4 vertices and every pair is connected.

🧮 Counting edges in Kₙ

• The excerpt asks: "What is the largest number of edges that a graph with 9 vertices can have?"
• Answer: at most "9 choose 2" edges, which is the number of ways to pick 2 vertices from 9.
• Generalization: Kₙ has "n choose 2" edges.
• Example: if you have 9 bus stops and want the maximum number of direct routes (at most one route between any two stops), the answer is the complete graph K₉.

🛤️ Path and cycle graphs

🛤️ Path graph Pₙ

Path graph Pₙ: the graph with n vertices and n − 1 edges that can be drawn so that all vertices and edges lie on a straight line.

• Vertices are arranged in a line; each vertex (except the two endpoints) is adjacent to exactly two others.
• The two endpoints have degree 1.
• Example: P₅ has 5 vertices and 4 edges.
• Real-world example from the excerpt: a bus route connecting Moby Arena → Corbett Hall → Library → CSU Transit Center → Engineering → MAX Station forms a path graph P₆.

🔄 Cycle graph Cₙ

Cycle graph Cₙ (for n ≥ 3): the graph with n edges and n vertices such that every vertex is adjacent to exactly two others.

• A cycle closes the path into a loop.
• Usually drawn as a circle.
• Example: C₆ has 6 vertices and 6 edges arranged in a circle.
• Real-world example: a bus loop connecting Weber → Engineering → MAX Station → back to Weber forms a cycle graph C₃.
• Don't confuse: a path has endpoints (degree 1); a cycle has no endpoints (every vertex has degree 2).

🍕 Bipartite graphs

🍕 What a bipartite graph is

Bipartite graph G = (V, E): V is the disjoint union of two sets L (left) and R (right), and every edge connects a vertex in L with a vertex in R; no edges exist between two vertices in L or between two vertices in R.

• The vertex set splits into two groups.
• Edges only cross between groups, never within a group.
• Example: three pizza parlors (left) and four student houses (right); an edge means "that house ordered from that parlor last month."
• No edges between two parlors (a parlor never orders from another parlor).
• No edges between two houses (a house never orders from another house).

🍕 Why bipartite graphs matter

• Many real-world problems involve matching two different types of objects:
  • Students with jobs
  • Pets with owners
  • Houses with pizza parlors
• The structure naturally models "two-sided" relationships.

🔗 Complete bipartite graph Bₙ,ₘ

Complete bipartite graph Bₙ,ₘ: the bipartite graph with n vertices on the left and m vertices on the right, such that every vertex on the left is adjacent to every vertex on the right.

• Every possible cross-group connection exists.
• Example: B₃,₄ models the situation where all 4 houses have ordered pizza from all 3 parlors within the last month.
• Counting:
  • Vertices: n + m total.
  • Edges: n × m (every left vertex connects to every right vertex).
• Special case: B₁,ₘ is called a "fan" (one vertex on the left connects to m vertices on the right).

📐 Degree of a vertex

📐 What degree means

Degree of v, denoted deg(v): the number of edges adjacent to v.

• It counts how many connections a vertex has.
• Example: at the CSU Transit Center bus stop, you can visit three other stops next → the degree of that vertex is 3.
• Example: at Horsetooth, there is only one route → the degree of that vertex is 1.

📐 Why degree matters

• Degree tells you how many choices you have at a given vertex.
• The excerpt introduces the question: "Given a set of n non-negative numbers, can this be the set of degrees of the vertices in a graph with n vertices?"
• This question explores whether a proposed degree sequence is realizable as a graph.

🧭 Overview

🧠 One-sentence thesis

The degree of a vertex counts how many edges connect to it, and the sum of all degrees in any graph must be even because each edge contributes exactly two to the total count.

📌 Key points (3–5)

• What degree measures: the number of edges adjacent to a given vertex.
• The fundamental constraint: the sum of all vertex degrees equals twice the number of edges, so the total is always even.
• Odd-degree vertices come in pairs: because the sum must be even, there must be an even number of vertices with odd degree.
• Common confusion: a vertex of degree 4 in a 5-vertex graph connects to all other vertices, so no vertex of degree 0 can exist in the same graph.
• Degree sequences: not every list of numbers can be the degrees of vertices in a valid graph—constraints like even sums and logical adjacency must hold.

🔢 What degree means

🔢 Definition and intuition

Degree of a vertex v, denoted deg(v): the number of edges adjacent to v.

• It counts how many connections a vertex has.
• Example: In the bus route graph, the CSU Transit Center connects to three other stops (Weber, Engineering, Horsetooth), so its degree is 3.
• Example: Horsetooth connects to only one other stop, so its degree is 1.

🚫 Don't confuse degree with other properties

• Degree is not the number of vertices in the graph; it is the number of edges touching one specific vertex.
• A vertex can have degree 0 (isolated, no edges) or degree equal to n−1 in an n-vertex graph (connected to every other vertex).

🧮 The fundamental theorem about degrees

🧮 Sum of degrees equals twice the number of edges

Theorem 8.3.6: The sum of the degrees of all vertices in a graph is twice the number of edges.

• Why: Each edge connects two vertices, so when you add up all degrees, every edge is counted once at each of its two endpoints.
• Example: A graph with 6 edges has total degree sum = 2 × 6 = 12.
• Consequence: The sum of all vertex degrees must be an even number.

🎭 Odd-degree vertices must come in pairs

• Because the total degree sum is even, the number of vertices with odd degree must also be even.
• Example: You cannot have exactly one vertex of odd degree, or exactly three—only 0, 2, 4, 6, etc.
• Two proofs given:
  1. Direct from the theorem: sum is even, so odd-degree vertices must pair up to contribute an even total.
  2. Induction on edges: adding one edge changes degrees of two vertices; in all three cases (even+even, even+odd, odd+odd), the parity of the count of odd-degree vertices stays even.

🧩 Valid and invalid degree sequences

✅ Valid examples

Vertices	Degrees	Valid?	Why
5	1, 2, 2, 2, 3	Yes	Sum = 10 (even); no logical conflict
9	1, 2, 2, 2, 3, 3, 3, 4, 4	Yes	Figure 8.2 is an example

❌ Invalid examples and why

❌ Logical conflict: degree 0 and degree 4 in a 5-vertex graph

• Degrees: 0, 1, 2, 3, 4
• Why invalid: A vertex of degree 4 in a 5-vertex graph must connect to all other four vertices, so no vertex can have degree 0 (isolated).
• Don't confuse: the sum here is 10 (even), but the adjacency logic is impossible.

❌ Odd sum

• Degrees: 2, 2, 3, 3, 3 (5 vertices)
• Sum: 2 + 2 + 3 + 3 + 3 = 13 (odd)
• Why invalid: Theorem 8.3.6 requires the sum to be even.

🧪 How to check a degree sequence

1. Check the sum: Is it even? If not, invalid.
2. Check odd-degree count: Is the number of odd degrees even? If not, invalid.
3. Check logical conflicts: Does a high-degree vertex force connections that contradict a low-degree vertex?
4. Try to draw it: Sometimes the only way to confirm is to attempt construction.

🔗 Degree in special graphs

🔗 Complete bipartite graph B(n,m)

• Left side (n vertices): each connects to all m vertices on the right, so degree = m.
• Right side (m vertices): each connects to all n vertices on the left, so degree = n.
• Example: In B(3,4), the 3 left vertices each have degree 4, and the 4 right vertices each have degree 3.

🔗 Fan graph B(1,m)

• One vertex on the left connects to m vertices on the right.
• The single left vertex has degree m; each of the m right vertices has degree 1.
• Called a "fan" because it looks like a central hub with spokes radiating out.

🧭 Overview

🧠 One-sentence thesis

A subgraph is formed by taking a subset of a graph's vertices and edges, and this concept allows us to study how graphs change when parts are removed or isolated.

📌 Key points (3–5)

• What a subgraph is: a graph whose vertices and edges are all contained in the original graph.
• How to identify valid subgraphs: both the vertex set and edge set must be subsets of the original; missing a vertex for an edge or adding new edges disqualifies it.
• Spanning subgraphs: special subgraphs that keep all vertices from the original graph.
• Common confusion: an object with edges but missing their endpoint vertices is not a subgraph—it's not even a valid graph.
• Boundary cases: every graph is a subgraph of itself, and the empty graph (no vertices or edges) is a subgraph of any graph.

🏗️ What makes a subgraph

🏗️ Formal definition

Subgraph: Let G = (V, E) be a graph. A graph G′ = (V′, E′) is a subgraph of G if V′ ⊆ V and E′ ⊆ E.

• The vertex set V′ must be a subset of the original vertex set V.
• The edge set E′ must be a subset of the original edge set E.
• Both conditions must hold simultaneously.

🚫 What disqualifies a subgraph

The excerpt provides clear examples of invalid subgraphs:

Case	Why it fails	Example from excerpt
Missing vertex for an edge	Not even a valid graph	Object (c): has an edge but missing one of its endpoint vertices
Extra edge not in original	Contains elements outside the original edge set	Graph (d): contains an edge that is not in G

• Don't confuse: removing an edge is allowed, but you cannot have an edge whose endpoints are not both present in your vertex subset.
• Example: if you keep an edge {a, b}, you must keep both vertices a and b.

🌉 Special types and examples

🌉 Spanning subgraphs

Spanning subgraph: a subgraph that contains all of the vertices of G.

• "Spanning" means the vertex set is complete (V′ = V), but the edge set can be any subset (E′ ⊆ E).
• Example from the bus route scenario: when a single bus route closes but all stations remain open, the result is a spanning subgraph with one edge removed.

🚌 Real-world scenario: bus routes

The excerpt uses a bus route graph to illustrate subgraph formation:

• First change: one bus route (edge) closes → spanning subgraph with one fewer edge.
• Second change: a station (vertex) closes → three adjacent routes must also close → subgraph with one fewer vertex and three fewer edges.
• This shows how removing a vertex forces removal of all its incident edges.

🔢 Boundary cases and structure

🔢 Trivial subgraphs

• The graph itself: every graph G is always a subgraph of itself (V′ = V, E′ = E).
• The empty graph: the graph with no vertices and no edges is a subgraph of any graph.
• These are the two extremes of the subgraph relationship.

📊 Subgraph enumeration

The excerpt shows a complete enumeration for a small graph (three vertices, one edge):

• All possible subgraphs are drawn in a hierarchy.
• Each connection represents removing a single vertex or edge.
• This illustrates that even small graphs can have many subgraphs.
• The exercises hint at systematic counting: count subgraphs with 0 vertices, then 1 vertex, then 2 vertices, etc.

🏷️ Labeled vs unlabeled

• When vertices are labeled (e.g., a, b, c), subgraphs with the same structure but different labels are considered different.
• Example: a subgraph with vertices {a, b} and edge {a, b} is different from one with vertices {b, c} and edge {b, c}, even though both have two vertices and one edge.