Session 07-03 - Combinatorics & Counting

Section 07: Probability & Statistics

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz - 10 Minutes

Quick Review from Session 07-02

Test your understanding of Basic Probability

  1. If \(P(A) = 0.4\) and \(P(B) = 0.3\) and A and B are independent, find \(P(A \cap B)\).

  2. Use the complement rule: If \(P(\text{rain}) = 0.25\), find \(P(\text{no rain})\).

  3. If \(P(A) = 0.5\), \(P(B) = 0.4\), and \(P(A \cap B) = 0.2\), find \(P(A \cup B)\).

  4. Are events A and B from question 3 independent? Why or why not?

Homework Discussion - 12 Minutes

Your Questions from Session 07-02

Let’s clarify notation before counting gets harder.

  • Independent vs mutually exclusive events
  • Complement strategy for “at least one”
  • Translating set notation into plain language

Learning Objectives

What You’ll Master Today

  • Apply the fundamental counting principle for sequential events
  • Calculate permutations: \(P(n,r) = \frac{n!}{(n-r)!}\)
  • Calculate combinations: \(C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}\)
  • Distinguish when order matters vs. doesn’t matter
  • Connect counting to probability calculations

Combinatorics is essential for calculating probabilities on the exam!

Part A: Fundamental Counting Principle

Sequential Choices

If you make sequential choices with \(n_1, n_2, ..., n_k\) options:

\[\text{Total possibilities} = n_1 \times n_2 \times ... \times n_k\]

Example: Creating an outfit

  • 4 shirts
  • 3 pairs of pants
  • 2 pairs of shoes

Total outfits: \(4 \times 3 \times 2 = 24\)

License Plate Example

A license plate has:

  • 3 letters (A-Z, 26 options each)
  • 4 digits (0-9, 10 options each)

Question: How many different plates are possible?

With repetition allowed: \[26 \times 26 \times 26 \times 10 \times 10 \times 10 \times 10 = 26^3 \times 10^4 = 175,760,000\]

Without repetition: \[26 \times 25 \times 24 \times 10 \times 9 \times 8 \times 7 = 78,624,000\]

Factorial Notation

Definition: Factorial

\[n! = n \times (n-1) \times (n-2) \times \dots \times 1\]

Special cases: \(0! = 1\) and \(1! = 1\)

Examples:

  • \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
  • \(3! = 3 \times 2 \times 1 = 6\)
  • \(10! = 3,628,800\)

Part B: Permutations

When Order Matters

Definition: Permutation

A permutation is an arrangement of objects where order matters.

\[P(n,r) = \frac{n!}{(n-r)!}\]

= Number of ways to arrange \(r\) objects from \(n\) distinct objects

Example: How many ways can 3 people win gold, silver, and bronze from 8 competitors?

\[P(8,3) = \frac{8!}{5!} = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336\]

Permutation Examples

Example 1: Arrange all letters in “MATH”

\[P(4,4) = 4! = 24 \text{ arrangements}\]

Example 2: How many 4-digit PINs with no repeated digits?

\[P(10,4) = \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5,040\]

Example 3: A president, VP, and treasurer from 12 candidates?

\[P(12,3) = 12 \times 11 \times 10 = 1,320\]

Part C: Combinations

When Order Doesn’t Matter

Definition: Combination

A combination is a selection of objects where order doesn’t matter.

\[C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}\]

= Number of ways to choose \(r\) objects from \(n\) distinct objects

Insight: Each combination corresponds to \(r!\) permutations

\[C(n,r) = \frac{P(n,r)}{r!}\]

Combination Examples

Example 1: Choose 3 students from 10 for a committee

\[C(10,3) = \binom{10}{3} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\]

Example 2: Choose 5 cards from a 52-card deck

\[C(52,5) = \binom{52}{5} = \frac{52!}{5! \cdot 47!} = \frac{52 \times 51 \times 50 \times 49 \times 48}{120} = 2,598,960\]

Decision Tree

If order matters, use permutations. If not, use combinations.

Quick Check - 6 Minutes

Decide the Method First

Work individually

For each situation, write either permutation or combination and compute:

  1. Choose 2 class representatives from 11 students.
  2. Assign president and vice president from 11 students.
  3. Number of 3-digit PINs from digits 0-9 (repetition allowed).

Break - 10 Minutes

Part D: Combinatorics in Probability

Connecting Counting to Probability

For equally likely outcomes:

\[P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\]

  • Counting helps us find both numbers!
  • Let’s look at some examples.

Question: Probability of being dealt exactly 2 hearts in a 5-card hand?

Example: Card Probability

Favorable: Choose 2 hearts from 13, AND 3 non-hearts from 39

\[\binom{13}{2} \times \binom{39}{3} = 78 \times 9,139 = 712,842\]

Total outcomes: Choose any 5 cards from 52

\[\binom{52}{5} = 2,598,960\]

Probability: \(P(\text{exactly 2 hearts}) = \frac{712,842}{2,598,960} \approx 0.274\)

Example: Committee Selection

A committee of 4 must be chosen from 6 men and 5 women.

Question: Probability that the committee has exactly 2 women?

  • Total ways to form committee: \(\binom{11}{4} = 330\)
  • Ways to get exactly 2 women: \(\binom{5}{2} \times \binom{6}{2} = 10 \times 15 = 150\)
  • Probability: \(P(\text{exactly 2 women}) = \frac{150}{330} = \frac{5}{11} \approx 0.455\)

Do you see a pattern here?

Example: Quality Control

Box contains 20 items: 4 are defective. We select 3 items randomly:

  1. \(P(\text{none defective})\)

\[P = \frac{\binom{16}{3}}{\binom{20}{3}} = \frac{560}{1140} = \frac{14}{28.5} \approx 0.491\]

  1. \(P(\text{at least one defective})\)

\[P = 1 - P(\text{none}) = 1 - 0.491 = 0.509\]

Part E: Pascal’s Triangle

A Shortcut for Binomial Coefficients

You’ve been computing \(\binom{n}{r}\) all session — for card hands, committees, and defective items. Pascal’s triangle arranges all these values in a pattern that makes them easy to look up and verify.

Pascal’s Triangle

The recursive property — each entry is the sum of the two above it:

\[\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}\]

Why this matters: Instead of computing \(\binom{5}{2} = \frac{5!}{2!\,3!}\) from scratch, find row \(n=5\) and look up the entry for \(r=2\) → 10. This saves time when verifying combination calculations.

Or you use your calculator or R to compute \(\binom{n}{r}\).

Guided Practice - 20 Minutes

Practice Problems

Work in pairs

Problem 1: A password must have 3 letters followed by 2 digits. a) How many passwords are possible (with repetition)? b) How many if no repetition is allowed?

Problem 2: From 8 candidates, we select a president, VP, secretary, and treasurer. How many ways?

Problem 3: A lottery requires choosing 6 numbers from 49. How many possible combinations?

Chained Exam Mini-Problem

Work individually, then compare

A committee of 4 is chosen from 10 people, including 3 finance majors.

  1. Compute total possible committees.
  2. Compute committees with exactly 2 finance majors.
  3. Use (a) and (b) to compute the probability of exactly 2 finance majors.

Training Tasks I

Work individually

An ice cream shop offers 12 flavors. You order a bowl with 3 scoops.

  1. How many bowls are possible if you choose 3 different flavors and the order of scoops doesn’t matter?
  2. Your friend orders a cone where the stacking order matters. How many cones with 3 different flavors are possible?
  3. What is the probability that a randomly chosen bowl (from part a) contains your two favorite flavors?

Training Tasks II

Work individually

A class has 15 students: 9 from Business and 6 from Economics. The professor randomly forms a project team of 5 students.

  1. How many possible teams are there?
  2. How many teams have exactly 3 Business and 2 Economics students?
  3. What is the probability that the team has exactly 3 Business and 2 Economics students?

Coffee Break - 10 Minutes

Collaborative Problem-Solving - 20 Minutes

Group Challenge: Event Planning

Think individually, then work in groups

An event team must choose from 9 volunteers:

  1. A chairperson, vice-chairperson, and treasurer.
  2. A separate 4-person logistics team.
  3. The probability that exactly 2 of 5 selected panelists are international students, if 7 of 18 applicants are international.

For each part, explain clearly whether order matters and why.

Counting Principle: Travel Routes

Work in pairs

A sales representative must visit 3 cities. She can travel:

  • From her office to City A by 4 routes
  • From City A to City B by 3 routes
  • From City B to City C by 2 routes
  1. How many different travel plans are possible?
  2. If she must also return from City C to her office and there are 3 return routes, how many round trips are possible?
  3. If one route from office to City A is closed, how many travel plans remain for part (a)?

Permutations vs. Combinations: Bookshelf

Work in pairs

A student has 8 different textbooks.

  1. How many ways can she arrange all 8 books on a shelf?
  2. How many ways can she choose 4 books to take on a trip?
  3. She wants to stack her 4 chosen books in a pile on her desk. How many different stacks are possible?
  4. Explain why (b) and (c) give different answers.

Probability with Combinatorics: Raffle Draw

Work individually, then compare

A company raffle has 30 tickets. You bought 4 tickets. Three winning tickets are drawn at random.

  1. How many ways can the 3 winning tickets be drawn?
  2. How many ways can all 3 winners come from your tickets?
  3. What is the probability that you win all 3 prizes?
  4. What is the probability that you win exactly 1 prize?

Mixed Problem: Digital Security

Work in groups

A company assigns 6-character access codes using digits 0-9 and uppercase letters A-Z (36 characters total).

  1. How many codes are possible if repetition is allowed?
  2. How many codes are possible if no character may repeat?
  3. A valid code must contain exactly 2 digits and 4 letters (no repetition, order matters). How many valid codes are there?
  4. If a code is generated at random (under the rules of part c), what is the probability that the two digits are adjacent?

Complement Strategy: Defective Parts

Work individually

A shipment contains 25 items, of which 5 are defective. A quality inspector selects 4 items at random.

  1. What is the probability that none of the selected items are defective?
  2. What is the probability that at least one selected item is defective?
  3. What is the probability that exactly 2 selected items are defective?

Final Assessment - 5 Minutes

Exit Ticket

Work individually

  1. How many 4-letter codes can be formed from 26 letters with repetition?
  2. How many ways to choose 3 students from 12?
  3. Which uses combinations: “arrange” or “choose”?

Wrap-Up & Key Takeaways

Today’s Essential Concepts

  • Counting principle: Multiply choices for sequential events
  • Factorial: \(n! = n \times (n-1) \times ... \times 1\)
  • Permutations: Order matters - \(P(n,r) = \frac{n!}{(n-r)!}\)
  • Combinations: Order doesn’t matter - \(C(n,r) = \frac{n!}{r!(n-r)!}\)
  • Key question: Does the order of selection matter?

Next Session Preview

Coming Up: Conditional Probability

  • Conditional probability: \(P(A|B)\)
  • Multiplication rule
  • Tree diagrams
  • Independence revisited

Homework

Complete Tasks 07-03:

  • Practice permutation and combination calculations
  • Solve probability problems using combinatorics