Permutations and Combinations

⏱ 10 min✏️ Quiz at the end

Introduction

Counting problems appear everywhere in mathematics and everyday life: how many ways can you arrange books on a shelf? How many different pizza combinations are possible? How many lottery ticket combinations exist? Two powerful tools β€” permutations and combinations β€” let us answer these questions without listing every possibility.

The key distinction to learn is simple:

  • Permutation β€” the order of items matters (arrangements)
  • Combination β€” the order does not matter (selections)

The Fundamental Counting Principle

Before exploring permutations and combinations, it helps to understand the Fundamental Counting Principle: if one task can be done in m ways and a second independent task can be done in n ways, then both tasks together can be done in m Γ— n ways.

Example: A restaurant offers 4 starters and 6 mains. The number of two-course meals possible is 4 Γ— 6 = 24.

This principle extends to any number of tasks: just multiply the number of choices at each stage.

Factorials

The factorial of a positive integer n, written n!, is the product of all positive integers from n down to 1.

n! = n Γ— (nβˆ’1) Γ— (nβˆ’2) Γ— ... Γ— 2 Γ— 1

Examples:

  • 3! = 3 Γ— 2 Γ— 1 = 6
  • 4! = 4 Γ— 3 Γ— 2 Γ— 1 = 24
  • 5! = 5 Γ— 4 Γ— 3 Γ— 2 Γ— 1 = 120

By definition, 0! = 1 (this might seem odd, but it makes the formulas below work out correctly).

Factorials grow very quickly: 10! = 3,628,800.

Permutations β€” When Order Matters

A permutation is an arrangement of items where the order in which items are placed or chosen matters.

The number of ways to arrange r items chosen from n distinct items is:

P(n, r) = n! / (n βˆ’ r)!

Example: How many ways can you arrange 3 letters chosen from the 5 letters {A, B, C, D, E}?

P(5, 3) = 5! / (5 βˆ’ 3)! = 120 / 2 = 60

You can also think about it step by step using the Fundamental Counting Principle:

  • 1st position: 5 choices
  • 2nd position: 4 remaining choices
  • 3rd position: 3 remaining choices
  • Total: 5 Γ— 4 Γ— 3 = 60 βœ“

Permutations of all n items: if you arrange all n items, the formula simplifies to just n! (since (n βˆ’ n)! = 0! = 1).

SituationFormulaExample
Arrange all n itemsn!4 books in a row: 4! = 24
Choose and arrange r from nP(n, r) = n!/(nβˆ’r)!Gold/silver/bronze from 8: P(8,3) = 336

Combinations β€” When Order Does Not Matter

A combination is a selection of items where order is irrelevant β€” only the group chosen matters.

Because order does not matter, every group of r items that was counted multiple times in permutations is now counted only once. We divide by r! to remove duplicate orderings:

C(n, r) = n! / (r! Γ— (n βˆ’ r)!)

This is also written as nCr or "n choose r" and sometimes shown as a binomial coefficient.

Example: How many 3-person committees can be chosen from 5 people?

C(5, 3) = 5! / (3! Γ— 2!) = 120 / (6 Γ— 2) = 120 / 12 = 10

Notice that C(5, 3) = 10 is much smaller than P(5, 3) = 60 because the committee {Alice, Bob, Cara} is the same regardless of the order in which they were selected.

Symmetry property: C(n, r) = C(n, n βˆ’ r). Choosing 3 from 5 gives the same count as choosing 2 from 5 to leave out. This is useful as a calculation shortcut.

SituationFormulaExample
Select r from n, order mattersP(n,r) = n!/(nβˆ’r)!Podium finishes: P(10,3) = 720
Select r from n, order irrelevantC(n,r) = n!/(r!(nβˆ’r)!)Team selection: C(10,3) = 120

Permutation or Combination? How to Decide

Ask yourself: would swapping two of the chosen items create a different outcome?

  • Yes β†’ Permutation. A PIN code of 1-2-3-4 is different from 4-3-2-1, so it is a permutation.
  • No β†’ Combination. A pizza with ham and mushroom is the same as a pizza with mushroom and ham, so it is a combination.

More examples:

ScenarioTypeReason
Arranging 5 books on a shelfPermutationEach arrangement looks different
Choosing 4 friends to inviteCombinationThe group is the same regardless of who was invited first
Assigning President, VP, Secretary from 10 membersPermutationThe roles are distinct
Picking 3 flavours from 8 for a tasting boxCombinationThe set of flavours is what matters

Worked Examples

Example 1 β€” Permutation: 8 athletes compete. How many ways can gold, silver, and bronze be awarded?

P(8, 3) = 8! / (8 βˆ’ 3)! = 8! / 5! = (8 Γ— 7 Γ— 6 Γ— 5!) / 5! = 8 Γ— 7 Γ— 6 = 336

Example 2 β€” Combination: A lottery requires choosing 6 numbers from 1 to 49. How many possible tickets exist?

C(49, 6) = 49! / (6! Γ— 43!) = (49 Γ— 48 Γ— 47 Γ— 46 Γ— 45 Γ— 44) / (6 Γ— 5 Γ— 4 Γ— 3 Γ— 2 Γ— 1) = 13,983,816

This is why winning the lottery is so unlikely β€” there are nearly 14 million possible combinations.

Common Mistakes

  • Confusing permutation with combination. Always ask whether the order of selection changes the outcome.
  • Forgetting 0! = 1. This is needed for the formulas to work when r = n or r = 0.
  • Using P when only the group matters. Choosing a committee of 3 from 10 is C(10, 3) = 120, not P(10, 3) = 720.
  • Calculation errors with large factorials. Cancel common factors before multiplying to keep numbers manageable. For C(49,6), cancel 43! from the numerator and denominator first.

Quick Reference

Factorial:     n! = n Γ— (nβˆ’1) Γ— ... Γ— 1     (0! = 1)
Permutation:   P(n,r) = n! / (nβˆ’r)!
Combination:   C(n,r) = n! / (r! Γ— (nβˆ’r)!)

Memory tip: Permutations are for Placements (order matters); Combinations are for Choosing a Crew (only membership matters).

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