Permutations and Combinations

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 — permutations and combinations — let us answer these questions without listing every possibility.

The key distinction to learn is simple:

• Permutation — the order of items matters (arrangements)
• Combination — 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 × n ways.

Example: A restaurant offers 4 starters and 6 mains. The number of two-course meals possible is 4 × 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 × (n−1) × (n−2) × ... × 2 × 1

Examples:

• 3! = 3 × 2 × 1 = 6
• 4! = 4 × 3 × 2 × 1 = 24
• 5! = 5 × 4 × 3 × 2 × 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 — 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 − 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 − 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 × 4 × 3 = 60 ✓

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

Situation	Formula	Example
Arrange all n items	n!	4 books in a row: 4! = 24
Choose and arrange r from n	P(n, r) = n!/(n−r)!	Gold/silver/bronze from 8: P(8,3) = 336

Combinations — When Order Does Not Matter

A combination is a selection of items where order is irrelevant — 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! × (n − 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! × 2!) = 120 / (6 × 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 − r). Choosing 3 from 5 gives the same count as choosing 2 from 5 to leave out. This is useful as a calculation shortcut.

Situation	Formula	Example
Select r from n, order matters	P(n,r) = n!/(n−r)!	Podium finishes: P(10,3) = 720
Select r from n, order irrelevant	C(n,r) = n!/(r!(n−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 → Permutation. A PIN code of 1-2-3-4 is different from 4-3-2-1, so it is a permutation.
• No → Combination. A pizza with ham and mushroom is the same as a pizza with mushroom and ham, so it is a combination.

More examples:

Scenario	Type	Reason
Arranging 5 books on a shelf	Permutation	Each arrangement looks different
Choosing 4 friends to invite	Combination	The group is the same regardless of who was invited first
Assigning President, VP, Secretary from 10 members	Permutation	The roles are distinct
Picking 3 flavours from 8 for a tasting box	Combination	The set of flavours is what matters

Worked Examples

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

P(8, 3) = 8! / (8 − 3)! = 8! / 5! = (8 × 7 × 6 × 5!) / 5! = 8 × 7 × 6 = 336

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

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

This is why winning the lottery is so unlikely — 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 × (n−1) × ... × 1     (0! = 1)
Permutation:   P(n,r) = n! / (n−r)!
Combination:   C(n,r) = n! / (r! × (n−r)!)

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