Counting Principles

TipLearning Objectives

By the end of this lesson, you’ll be able to:

  • Use the Fundamental Counting Principle to count outcomes.
  • Compute permutations and combinations.
  • Distinguish ordered vs. unordered selections.
  • Apply counting to probability problems.

Key Ideas

Fundamental Counting Principle

If event A has \(m\) choices and event B has \(n\) choices: \[ m \times n \text{ total outcomes} \]

Permutations (order matters)

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

Combinations (order does NOT matter)

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

Common Problem Types

Sequential Choices

Multiply number of choices at each step.

Example:
3 shirts × 2 pants = 6 outfits.

Arrangements (Permutations)

Order matters.

Example:
Arrange 4 students in a line → \(4!=24\).

Choosing Groups (Combinations)

Order does NOT matter.

Example:
Choose 3 committee members out of 10 → \(\binom{10}{3}\).

Counting in Probability Problems

Probability = favorable outcomes ÷ total outcomes.

Example:
Ways to pick 2 red cards ÷ ways to pick 2 cards total.

Avoiding Overcounting

Use combinations when order would double-count.

Example:
Picking Alice + Bob = same as Bob + Alice.

Strategies

  • First decide: Does order matter?
  • Use the counting principle for multi-step choices.
  • Use permutations for arrangements.
  • Use combinations for groups.
  • Simplify factorials early to avoid large numbers.

Worked Examples

Example 1

License plate uses 3 letters then 3 digits: \[ 26^3 \cdot 10^3 \]

Example 2

From 8 runners, choose top 3 finishers in order: \[ P(8,3)=\frac{8!}{5!}=336 \]

Example 3

From 10 students, choose 4 for a project team: \[ C(10,4)=210 \]

WarningCommon Mistakes
  • Using permutations when order doesn’t matter.
  • Forgetting to multiply choices for multi-step events.
  • Overcounting due to order duplication.
  • Forgetting factorial simplifications.

Practice Problems

  1. 4 shirts, 3 pants → how many outfits?
  2. Arrange 5 books on a shelf.
  3. Choose 2 students out of 12.
  4. From digits 1–5, how many 3-digit codes with no repeats?
  1. \(4 \cdot 3 = 12\)
  2. \(5! = 120\)
  3. \(\binom{12}{2}=66\)
  4. \(5 \cdot 4 \cdot 3 = 60\)

Summary

  • Counting principle multiplies choices.
  • Permutations: order matters.
  • Combinations: order does not matter.
  • Ask “Does order matter?” first.
  • Use combinations to avoid double-counting.
  • Break complex problems into sequential choices.