L-PR-3: Counting Techniques

Module 2 · Probability Foundations

Section 1: Introduction

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

A smartphone lock screen with a 6-digit PIN has 1,000,000 possible codes. A 6-letter password from the 26 letters of the alphabet — no repeats — has 165,765,600 possibilities. Why the enormous difference?

In L-PR-1, you counted outcomes by listing them. That works for small sample spaces. This lesson gives you three systematic methods that handle problems too large to list — and they plug directly into the probability formula you already know.

In L-PR-1, — but you found and by listing. Today you build the machinery to count and without listing a single outcome.

After this lesson, you will be able to:

Apply the Fundamental Counting Principle to count outcomes for multi-stage processes.
Compute factorials and use them in the permutation and combination formulas.
Calculate permutations and combinations for problems with ordered and unordered selections.
Identify which counting method applies and use it to compute probabilities over equally-likely sample spaces.

Section 2: Prerequisites

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

What you need from L-PR-1

Equally-likely outcomes and (L-PR-1, Core Concepts C1–C2): Every counting technique in this lesson feeds directly into this formula. We will count and systematically — but the formula connecting them to probability is the one you already know.
Reading a sample space (L-PR-1, Core Concepts C2–C3): You listed sample spaces for small experiments. Today we replace that listing with formulas. Understanding what a sample space is — and why it must be complete — remains essential.
The complement rule (L-PR-1, Core Concepts C5): . In Section 9 (Challenge Problems), you will use the complement to avoid counting a large favorable set directly. The rule is the same; only the counting method is new.

Retrieval Checkpoint

A bag contains 5 red marbles and 3 blue marbles. One marble is drawn at random. What is ?

Also confirm you can recall:

I can identify n(S), the total number of equally-likely outcomes in a sample space. I know that

and can apply the complement rule. I understand that

requires all outcomes to be equally likely.

Success Factor:

From listing to counting. In L-PR-1, finding meant listing outcomes or using a simple sample space. Today we unlock three methods — the Fundamental Counting Principle, permutations, and combinations — that count thousands or millions of outcomes without listing a single one, then feed those counts directly into .

Section 3: Core Concepts

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Navigation Guide — 6 Concepts

C1: Fundamental Counting Principle — “How many complete sequences are possible?” Multiply the available choices stage by stage, using the correct count at each stage.
C2: Factorials — the building block for permutation and combination formulas.
C3: Permutations — “How many ordered arrangements?” Use when the order of selection produces a different outcome.
C4: Combinations — “How many unordered selections?” Use when two selections with the same objects in different order count the same.
C5: Permutation vs. Combination decision — the single most important judgment call in this lesson.
C6: Counting-based probability — feeding counting results back into .

C1 — Fundamental Counting Principle

Imagine a restaurant with 3 starters, 5 main courses, and 4 desserts. How many distinct three-course meals can you order?

You have 3 choices for the starter. For each starter, 5 choices for the main. For each (starter, main) pair, 4 choices for dessert. Total meals:

This is the Fundamental Counting Principle (FCP): multiply the number of available choices at each stage.

Fundamental Counting Principle

Suppose a sequential process has stages. If every partial sequence that reaches stage has exactly available continuations, then the total number of complete sequences is:

The factors may differ from one stage to the next. Probabilistic independence is not required: without replacement, for example, the available counts can be . If different branches at the same stage have different continuation counts, count those branches separately and add their totals.

Mini-example (2 stages): A student can take a morning class (4 options) and an afternoon class (6 options). How many distinct (morning, afternoon) pairs are possible?

pairs. Each morning choice can be followed by any of the 6 afternoon options.

A tree diagram makes the multiplication visible. With 2 morning choices and 3 afternoon choices, every Stage-1 branch fans into 3 Stage-2 branches — giving complete paths:

Figure 1: Each Stage-1 branch fans into 3 Stage-2 branches, so the six complete paths are counted by

, not

Adjust the choices at each stage and watch the tree grow multiplicatively. Try setting Stage 1 to 4 and Stage 2 to 3 — notice how every Stage-1 branch fans into exactly 3 Stage-2 branches, giving distinct paths. Then switch on No repeats: each chosen item leaves the pool, so the factors shrink from stage to stage. The product rule still applies; only the available-choice counts change.

Stage 1 =

Stage 2 =

Running product

Figure 2: The interactive tree shows one complete path for every sequence of stage choices, including shrinking stage counts when repeats are forbidden.

“No repeats means I cannot multiply.” You still use the product rule — you just update each factor. A 4-digit PIN with distinct digits has 10 choices, then 9, then 8, then 7, so the count is , not . The same falling product is written compactly as the permutation in C3.

Keep the vocabularies separate: with repetition means a used item remains available, while without repetition (or without replacement) means it does not. Those conditions determine the stage counts; they are not definitions of probabilistic independence.

C2 — Factorials

How many ways can you arrange 3 books (A, B, C) in a row?

Position 1: 3 choices
Position 2: 2 remaining choices
Position 3: 1 remaining choice

Total: . List them to verify: ABC, ACB, BAC, BCA, CAB, CBA — exactly 6.

This product occurs so often it gets its own symbol.

Factorial

For any positive integer :

Convention: .

Factorials count the number of ways to arrange all distinct objects in a sequence. They grow rapidly: 5! = 120, , .

Why : The permutation formula (C3) requires to equal — the number of ways to arrange all objects. This forces . Also: there is exactly one way to arrange zero objects (do nothing).

See the cancellation: Expand into individual multiplied terms. Click “Show cancellation” to watch the denominator strike through the matching tail of the numerator, leaving only the highlighted terms — that is exactly how the permutation formula works.

n =

r =

Figure 3: The factorial strip expands

and highlights the uncancelled factors that form a permutation count.

"" is the most common factorial error. It makes undefined or zero — but there is exactly one way to choose nothing from a set (the empty selection). Always: .

C3 — Permutations: Ordered Arrangements

In a club of 8 members, how many ways can we elect a president, vice-president, and secretary (3 distinct roles)?

Apply the FCP with shrinking stage counts: 8 choices for president, 7 for VP (one member used), and 6 for secretary.

This is — a permutation. We can write this more compactly using factorials:

Permutation

A permutation selects objects from distinct objects and arranges them in order. We write :

Use a permutation when order matters — a different arrangement of the same objects is a different outcome.

Special case: (arranging all objects).

Derivation from FCP: For position 1, choices. Position 2: . Position 3: . … Position : . Product = . Multiplying numerator and denominator by closes the form:

Use only when order changes the outcome. If you are selecting a 3-person team from 8 (not assigning roles), the order of selection does not matter — {Ann, Bob, Carl} and {Carl, Ann, Bob} are the same team. That is a combination (C4). Using ₈P₃ = 336 when the answer is ₈C₃ = 56 overcounts by .

Build a permutation: Click an object from the pool to place it in the next position. Watch the choice count above each slot decrease as the pool shrinks — that shrinking count is exactly where the formula comes from.

Available pool — click to place

Your arrangement

Running product

Figure 4: Building an ordered selection makes the available pool shrink by one after each position is filled.

C4 — Combinations: Unordered Selections

From 8 candidates, how many 3-person committees can be formed (no roles assigned)?

Each group of 3 candidates, regardless of the order we list their names, is the same committee. The permutation count of 336 overcounts each committee: every committee of 3 people can be arranged in orders — (A,B,C), (A,C,B), (B,A,C), (B,C,A), (C,A,B), (C,B,A) — all referring to the same committee.

Combination

A combination selects objects from distinct objects without regard to order. We write or :

Use a combination when order does not matter — two selections with the same objects in a different order count as the same outcome.

Properties:

(exactly one way to choose nothing, or choose everything)
Symmetry: (choosing to include is the same as choosing to exclude)

Symmetry as a shortcut: . When is large, use the symmetry to work with the smaller number.

Forward pointer to L-PR-5: In L-PR-5 (Binomial Distribution), the combination formula appears inside the binomial probability: . The counts the number of ways exactly successes can occur in trials — the formula you just learned is the key ingredient.

See the symmetry: The same split of five objects can be read two ways. Flip the reading below — choosing 2 to include is choosing 3 to exclude, so the two counts must be equal.

Read the split as:

{A, B} go in → C(5,2) = 10 ways to choose the 2 included.

Figure 5: The same split can be described by the objects included or excluded, illustrating

Explore permutations and combinations interactively:

Select a value of below. The left panel shows all ordered arrangements; the right panel shows all unordered selections. Matching tints mark the orderings that contain the same objects — switch the group view to Grouped to watch them collapse into a single combination when order is ignored.

Ordered — ₄P_r 12

12 ÷ 2! = 6

Unordered — ₄C_r 6

Group view:

Figure 6: Ordered arrangements collapse into unordered groups when sequences containing the same selected objects are treated as equivalent.

C5 — Permutation vs. Combination: The Decision

The core question is: does changing the order of the selected items give a different outcome?

Yes → Permutation. “Being first is different from being second.”
No → Combination. “Only the membership of the group matters.”

Scenario	Order matters?	Method
Elect president, VP, secretary from 12 members	Yes — different roles	Permutation
Form a 3-person committee from 12 members	No — same committee regardless of listing order	Combination
Arrange 5 songs on a playlist	Yes — different order is a different playlist	Permutation
Choose 3 toppings from a menu of 8	No — same toppings in any order	Combination

Signal words:

→ Permutation: “arrangement,” “ranking,” “ordered list,” “schedule,” “sequence,” “1st, 2nd, 3rd,” “code or password (when the problem states that order matters)”
→ Combination: “group,” “team,” “committee,” “set,” “selection,” “sample,” “subset”

“Combination lock” is a misleading counting name. A lock where 3-7-4 is different from 4-7-3 uses an ordered code, not an unordered combination. If digits may repeat, count codes with the FCP; if they may not, use a permutation. The everyday English use of “combination” does not match the mathematical meaning. In mathematics, a combination is an unordered selection.

Use this flowchart to classify the counting problems covered in this lesson. Expand the examples to see the formula applied to the scenarios from this lesson.

Repetition changes the factors, not the order question.
Decide first whether swapping two selected items changes the outcome.

Does swapping two selected items
produce a different outcome?

YES — swapping changes the outcome

Order matters

Repetition allowed → FCP

n × n × ⋯ × n = n^r

When each of the r distinct positions has the same n choices.

Repetition forbidden → Permutation

_nP_r = n! / (n − r)!

Each selection removes one item, so the stage counts shrink.

arrangement ranking ordered list schedule code (no repeats)

Electing a president, VP, and secretary from 12 — which branch? Check

Swapping president and VP gives a different outcome — order matters, so Permutation.
₁₂P₃ = 12 × 11 × 10 = 1320

NO — swapping does not change the outcome

Order does not matter

Repetition forbidden → Combination

C(n, r) = n! / [r! · (n − r)!]

Different orders of the same selected items collapse to one outcome.

Repetition allowed → Separate case

Unordered selection with repetition uses a different method and is outside this lesson.

group team committee set selection

Choosing a 3-person committee from 12 — which branch? Check

{Ann, Bob, Carl} and {Carl, Ann, Bob} are the same committee — order doesn't matter, so Combination.
C(12, 3) = 12! / (3! · 9!) = 220
Note: 220 = 1320 ÷ 3! — the same 12 · 11 · 10 selections, with each group's 6 orderings collapsed into one.

"Combination lock" is a misleading counting name: its code is ordered because 3‑7‑4 ≠ 4‑7‑3. Use the FCP if digits may repeat, or a permutation if they may not.

Figure 7: The decision flow separates sequential products from ordered permutations and unordered combinations.

C6 — Counting-Based Probability

When all outcomes in a sample space are equally likely, . Counting techniques give us both and when manual enumeration is not feasible.

The formula from L-PR-1 still applies — we just use counting tools to find the numbers:

Counting-Based Probability

When all outcomes are equally likely:

The numerator and denominator must count outcomes at the same level of detail. For example, do not count ordered sequences in the numerator and unordered groups in the denominator.

End-to-end example: From a class of 20 students (8 who are athletes, 12 who are not), 3 are chosen at random. Find .

Total outcomes: equally-likely committees.

Favorable outcomes: All 3 from the 8 athletes: .

Explore all four outcomes: The bar below divides all 1,140 equally-likely committees by the number of athletes included. Click any segment to see the combination formula and exact probability.

Click a segment to see the probability formula

Figure 8: All 1,140 three-person committees are partitioned by the number of athletes selected, making each favorable count part of the same sample space.

“The numerator and denominator must use the same named formula.” They must describe the same kind of equally-likely outcome, but different counting tools can be compatible. You might count a restricted numerator with a short stage product and the denominator with a permutation formula if both count ordered sequences. The real error is mixing ordered outcomes with unordered ones.

Counting by complement. When the favorable set is large but its complement is small, it is easier to count what you do not want and subtract: . For example, is easier as than by summing three direct cases. This strategy becomes essential whenever a problem uses “at least one” or “at least ” phrasing — you will use it in the Challenge Problems (Section 9) and again in L-PR-5.

The bar below makes the strategy visible: the “0 athletes” slice is the only thing you have to count — everything else is “at least one,” no case-by-case work required.

Figure 9: The single “no athletes” segment is the complement of all committees containing at least one athlete.

The opening hook — resolved. A 6-digit PIN uses 10 digits with repetition allowed, so every position has 10 choices and the FCP gives . A 6-letter password with no repeated letters has shrinking stage counts: . The enormous difference comes from whether used symbols remain available, not from whether multiplication is allowed.

Section 4: Worked Examples

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Example 1 — Fully Worked: FCP and Permutation

Part A: A restaurant has 4 appetizers and 6 main courses. How many distinct (appetizer, main course) pairs are possible?

Part B: In how many ways can a president, vice-president, and secretary be elected from a club of 12 members?

Part A — FCP:

I see two stages: choose an appetizer (4 options), then choose a main course (6 options). Every appetizer can be followed by any of the 6 mains, so I multiply the stage counts.

Total pairs = .

Part B — Permutation:

First I ask: does order matter? Yes — being elected president is a different outcome from being elected vice-president, even if the same person is chosen. The three roles are distinct. So I use a permutation.

members, positions. I write out the product from the FCP derivation to check my answer:

Using the formula: ✓

I pause to check: is 1320 reasonable? There are 12 choices for president, then 11 for VP, then 10 for secretary — . Confirmed.

Example 2 — Prediction Checkpoint: Combinations

A committee of 4 people is chosen from 10 candidates. No roles are assigned — the 4 members are equal. Without calculating exactly, is closer to 200 or to 5040? Then compute the count and explain the size difference.

Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.

Show Solution

Compute:

The answer is close to 200 — much smaller than 5040, which is , the ordered version. Dividing by removes the overcounting from treating different orderings of the same four people as distinct.

Symmetry check:

Thus . Choosing four to include is equivalent to choosing six to exclude.

Example 3 — Details/Summary: Counting-Based Probability

A class of 30 students contains 12 who play varsity sports and 18 who do not. Five students are randomly selected to present projects. What is the probability that exactly 3 of the 5 selected are athletes?

Pause and commit. Decide what one equally-likely outcome is, then write expressions for the total and favorable counts on paper or in your notes. Open the solution afterward and compare your setup and method choice.

Show Solution

Step 1 — Identify the method. Order of selection doesn’t matter (3 athletes chosen from 12, 2 non-athletes chosen from 18, for a total committee of 5). Use combinations throughout.

Step 2 — Count total outcomes:

Step 3 — Count favorable outcomes: Choose 3 athletes from 12 and, for each such group, choose 2 non-athletes from 18. Multiply the two counts:

Step 4 — Compute the probability:

Check the method consistency: Both numerator and denominator used combinations — appropriate since we’re forming an unordered group.

Example 4 — Find the Error

A student is asked: “In how many ways can a 4-person project team be formed from 9 students?” The student writes:

Student’s work:

“The team needs 4 people. The order in which we pick them gives the team its structure, so I should use a permutation."

"There are 3024 possible teams.”

Identify and correct the error:

Pause and commit. Diagnose the student’s error and write a corrected calculation on paper or in your notes. Then open the solution and compare both your explanation and your count.

Show Solution

The error is using a permutation when the problem asks for a team (unordered group). A project team has no internal ranking — {Ann, Bob, Carl, Dana} and {Dana, Carl, Bob, Ann} are the same team. The permutation count of 3024 treats every ordering of the same 4 people as a separate team, overcounting each team by .

Correct calculation:

There are 126 possible teams — exactly , confirming the overcounting factor.

Key signal missed: The word “team” signals an unordered selection. Had the problem asked for a president, secretary, treasurer, and recorder, a permutation would be correct.

Section 5: Guided Practice

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Counting-Probability Pipeline (C4 + C5 + C6)

Work through the full decision-and-probability pipeline: classify the selection, count all outcomes, count the favorable outcomes, and compute the probability.

Problem 2 — Combination Calculation and Symmetry (C4)

Compute and verify the symmetry property . You may try the whole decision first or ask for the guided steps.

Problem 3 — Fundamental Counting Principle (C1)

Apply the product rule to count the outcomes of a multi-stage process.

Problem 4 — Permutation Calculation (C3)

Given a scenario where order matters, compute the permutation.

Problem 5 — Factorial Arrangements (C2)

Compute the number of ways to arrange all of a set of distinct objects.

Section 6: Independent Practice

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — FCP Probability (C1 + C6)

Count total and favorable outcomes using the FCP, then compute the probability.

Problem 2 — Permutation vs. Combination Classification (C3 + C4 + C5)

Classify each scenario and compute the result.

9 books are available; 4 are chosen for an ordered reading list (the order matters — book 1 is read first, book 4 last).

(a) Is this a permutation or combination? Why?

A 5-member jury is selected from 12 candidates. No roles are assigned — all jurors are equal.

(a) Is this a permutation or combination? Why?

7 runners compete. Distinct medals are awarded for 1st, 2nd, and 3rd place.

(a) Is this a permutation or combination? Why?

From 8 ice cream flavours, a customer picks 2 different scoops. The order of scoops on the cone does not change the dessert.

(a) Is this a permutation or combination? Why?

A password is 4 distinct letters from a pool of 10, and ABCD is not the same password as DCBA.

(a) Is this a permutation or combination? Why?

Problem 3 — Combination-Based Probability (C4 + C6)

From a group with a specific subgroup, find the probability that a random selection includes a given number from the subgroup.

Problem 4 — Find the Counting Error (C2 + C3 + C4 + C5)

A student’s counting calculation is shown below. Identify the specific error and determine the correct answer.

Problem 5 — Hiring Committee Synthesis (C4 + C5 + C6)

A company’s hiring committee is reviewing 15 candidates: 9 with industry experience, 6 without. The committee selects 6 candidates to interview (equally-likely random selection).

(a) How many ways can 6 candidates be selected from 15?

(b) How many selections include exactly 4 experienced candidates?

(c) What is ?

(d) Which method requires fewer mutually exclusive cases to sum by hand for : direct calculation or the complement?

Show Solution

(a)

(b) Choose 4 from the 9 experienced candidates AND 2 from the 6 inexperienced candidates: $₉ ₄ ₆ ₂$

(c)

(d) Direct computation requires summing — three terms. The complement requires summing four terms (exactly 0, 1, 2, 3 experienced). Direct is marginally simpler here.

Preview: In L-PR-5 (Binomial Distribution), you will encounter a systematic formula for this type of “exactly successes” problem.

Section 7: Mastery Check

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Question 1 — Feynman Test

Explain to a classmate why the Fundamental Counting Principle still works when choices are made without replacement, even though the factors shrink. Use a four-digit code with no repeated digits as your example, and distinguish the product rule, replacement, and probabilistic independence.

Compare with a model answer

The Fundamental Counting Principle says to multiply the number of available continuations at each stage. It does not require probabilistic independence, and the factors do not have to stay equal.

For a four-digit code with no repeated digits, the first position has 10 choices, the second has 9, the third has 8, and the fourth has 7:

Writing would be wrong because it keeps all ten digits available at every stage. “Without replacement” tells us to shrink the available-choice count; it does not tell us to stop multiplying. The same falling product is the permutation .

Question 2 — Apply

A student council has 8 members. A committee of 3 must be formed, and one specific member (the student body president) must always be included.

(a) How many such committees are possible?

(b) Now consider a different process: a 3-person committee is selected uniformly at random from all possible 3-person committees, with no member required in advance. What is the probability that the selected committee includes the president?

Solution

Part (a): The president is fixed on the committee. Choose the remaining 2 members from the 7 others:

Part (b): Total possible 3-person committees from 8 members: ₈C₃ = 56.

Alternative via complement: . So ✓

Question 3 — Error Analysis

A student is asked: “8 students are competing for 3 prizes — 1st, 2nd, and 3rd place. How many outcomes are possible?”

Student’s work: “Since we’re choosing 3 from 8, I use a combination: outcomes.”

What is the specific error?

Solution

The error is using a combination when a permutation is required. The three prizes (1st, 2nd, 3rd place) are distinct — awarding gold to Student A and silver to Student B is a different outcome from gold to B and silver to A.

Correct calculation:

There are 336 possible ranked outcomes, not 56. The combination ₈C₃ = 56 would be correct only if the 3 prizes were identical (e.g., 3 students each receive the same award with no ranking).

Self-check using the relationship: ✓. Permutations = Combinations × (arrangements within the group).

Self-Assessment

How confident do you feel about the material in this lesson?

Still confusedReady for the Boss Fight

Section 8: Boss Fight

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Choose your path. Both paths require mastery of counting techniques — one tests computational fluency, one tests conceptual judgment.

🔐 Path A — The Cryptographer

Build passcodes for a new security app and compute probabilities under varying constraints.

🏒 Path B — The Tournament Director

Resolve a dispute about counting methods in a sports tournament, then compute probabilities.

Path A — The Cryptographer

Pause and commit. On paper or in your notes, explain what changed between the no-repetition and repetition-allowed counts. Why did one product use shrinking factors while the other used the same factor at every stage? Then compare your explanation with the model reflection.

Show Model Reflection

The product rule applies in both cases. Without repetition, each chosen character leaves the pool, so the available counts shrink: , then , then , and so on. With repetition allowed, every character remains available, so every position has choices and the count is . Replacement determines the factors; it should not be used as a synonym for probabilistic independence.

Path B — The Tournament Director

A hockey tournament has 10 teams. Two directors disagree about counting podium outcomes.

Director A claims: “The number of possible gold-silver-bronze podium outcomes is .”

Director B says: “No — it’s .”

Task 1. Who is correct and why? Identify the specific error in the incorrect director’s reasoning.

Recall: gold, silver, and bronze are distinct awards — the order matters. Write a clear explanation of the error, naming the correct formula and its justification.

Task 2. After the tournament, a 3-team organizing committee is formed from all 10 teams. No team has a higher role than another — all three committee roles are equal.

How many distinct organizing committees are possible?

Is this a permutation or combination? Apply the appropriate formula.

Show Solutions — Tasks 1–2

Task 1: Director B is correct. Gold, silver, and bronze carry distinct ranks — awarding gold to Team X and silver to Team Y is a different outcome from gold to Y and silver to X. Director A’s error was applying , which counts unordered groups of 3 teams and ignores which team won which medal. Reconciliation: ✓.

Task 2: No roles assigned → unordered selection → combination.

Key signal words: When the order of selected items produces a different outcome (medals, roles, ranked positions) → permutation. When only the membership of a group matters (committees, teams with equal roles) → combination.

Pause and commit. In one or two sentences on paper or in your notes, contrast the ordered podium outcomes, unordered committees, and probability questions in this path. Name the decision that determines each count, then compare with the model reflection.

Show Model Reflection

The podium is an ordered outcome, so swapping teams between gold, silver, and bronze creates a new permutation. The organizing committee is unordered, so only membership matters and a combination is appropriate. For each probability, I first define one equally-likely outcome at the matching level of detail, then divide the favorable count by the total count.

Section 9: Challenge Problems

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Ready for more? These go beyond the lesson objectives — they are entirely optional.

Problem 1 — Algebraic Proof + Combinatorial Argument

(a) Prove algebraically that .

Which of the following correctly shows written out and simplified?

(b) Give a combinatorial word argument: “Choosing objects to include is the same as choosing objects to exclude.”

Verify with , : List all pairs of objects chosen from {A, B, C, D, E}, and for each pair, identify the complementary group of 3 excluded objects.

Which reasoning below correctly restates the word argument?

(c) Evaluate efficiently, without expanding .

Show Complete Solution

(a) Algebraic proof

(b) Combinatorial argument

Every choice of included objects uniquely determines the objects left out, and every excluded group uniquely determines its included complement. This one-to-one pairing proves that the two counts are equal.

For {A, B, C, D, E}, the ten included pairs and excluded triples are:

AB ↔ CDE
AC ↔ BDE
AD ↔ BCE
AE ↔ BCD
BC ↔ ADE
BD ↔ ACE
BE ↔ ACD
CD ↔ ABE
CE ↔ ABD
DE ↔ ABC

Thus .

(c) Efficient evaluation

Problem 2 — Card Probability

Find the probability that at least one card of a specified suit appears in a 5-card hand.

Problem 3 — Restricted Arrangements

Six friends — Alex, Blake, Cam, Dana, Evan, and Fran — are seated in a row of 6 chairs.

(b) How many arrangements have Alex and Blake sitting next to each other?

(c) How many arrangements have Alex and Blake NOT adjacent?

(d) What is when the seating is assigned randomly?

Show Complete Solution

(a) Total arrangements

(b) Alex and Blake adjacent

Treat Alex and Blake as one block. The block plus the other four people makes five objects to arrange:

Alex and Blake can appear inside their block in two orders:

(c) Alex and Blake not adjacent

Use the complement:

(d) Probability

Section 10: Solutions Reference

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Quick-Reference Formulas

Formula	Name	When to use
	Fundamental Counting Principle	Sequential stages with the correct available-choice count at each stage; probabilistic independence is not required
;	Factorial	Arranging all distinct objects
	Permutation	Ordered selection of from (order matters)
	Combination	Unordered selection of from (order irrelevant)
	Symmetry	Shortcut when is large; self-check
	Counting-based probability	Equally-likely outcomes; use FCP/perm/combo for both numerator and denominator

Decision rule: Ask “does swapping two selected items give a different outcome?” Yes → permutation. No → combination.

Solutions and Explanations

For full worked solutions to all Practice, Mastery, Boss Fight, and Challenge problems in this lesson, see the solutions page:

View Full Solutions