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?

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

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

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

Key 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

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

Permutation vs Combination Decision Tree

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!

Example: Card Probability

What is the probability of being dealt exactly 2 hearts in a 5-card hand?

Favorable outcomes: 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: What is the 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: \[P(\text{exactly 2 women}) = \frac{150}{330} = \frac{5}{11} \approx 0.455\]

Example: Quality Control

A box contains 20 items: 4 are defective. If 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

Binomial Coefficients Pattern

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

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?

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