Back to Chapter 1

Section 1.2: Methods of Enumeration

Master counting techniques: multiplication principle, permutations, combinations, and multinomial coefficients.

The Multiplication Principle

If an experiment has two parts where the first part has m outcomes and the second part has n outcomes, then the total number of outcomes is m × n.

Example: License Plates

A license plate has 3 letters followed by 4 digits. How many unique plates are possible?

26×26×26×10×10×10×10=175,760,000

Permutations

A permutation is an ordered arrangement. When order matters, we use permutations.

Factorial: n!

The number of ways to arrange all n distinct objects.

n! = n × (n-1) × (n-2) × ... × 2 × 1

Note: 0! = 1 by definition

P(n,r) or ₙPᵣ

The number of ways to select r objects from n where order matters.

P(n,r) = n! / (n-r)!

= n × (n-1) × ... × (n-r+1)

When to Use Permutations

  • • Arranging people in a line or seats
  • • Awarding 1st, 2nd, 3rd place (rankings)
  • • Creating passwords/codes where order matters
  • • Any scenario where ABC ≠ BAC ≠ CAB

Combinations

A combination is a selection where order does not matter. Also called "n choose r" or binomial coefficient.

Formula: C(n,r) or (n choose r)

C(n,r) = n! / [r! × (n-r)!]
= P(n,r) / r!

Key Properties

  • • C(n,r) = C(n, n-r) — symmetry
  • • C(n,0) = C(n,n) = 1
  • • C(n,1) = n
  • • Sum of row n = 2ⁿ

When to Use Combinations

  • • Selecting a committee (no positions)
  • • Choosing cards in poker/bridge hands
  • • Picking lottery numbers
  • • Any scenario where {A,B,C} = {C,A,B}

Sampling: With vs Without Replacement

ScenarioOrderedUnordered
With ReplacementC(n+r-1, r)
Without ReplacementP(n,r) = n!/(n-r)!C(n,r) = n!/[r!(n-r)!]

Exam Tip

Always ask yourself two questions: (1) Does order matter? (2) Can items be reused? This determines which formula to use.

Distinguishable Permutations & Multinomial Coefficients

When arranging n objects where some are identical, we divide by the factorials of the group sizes.

Multinomial Coefficient

n! / (n₁! × n₂! × ... × nₖ!)
where n = n₁ + n₂ + ... + nₖ

Example: MISSISSIPPI

How many ways to arrange the letters in "MISSISSIPPI"?

11 letters: M(1), I(4), S(4), P(2)

11! / (1! × 4! × 4! × 2!) = 34,650 arrangements

Quick Reference: Key Formulas

Factorial:n! = n × (n-1) × ... × 1
Permutation:P(n,r) = n!/(n-r)!
Combination:C(n,r) = n!/[r!(n-r)!]
Multinomial:n!/(n₁!×n₂!×...×nₖ!)