Permutations vs Combinations: What’s the Difference?

A permutation counts arrangements where order matters. A combination counts selections where order doesn’t matter. That’s the whole distinction — but knowing which one to use is the trick.

Quick answer

If order matters, use a permutation: P(n, r) = n! / (n – r)!. If order doesn’t matter, use a combination: C(n, r) = n! / [r!(n – r)!]. The difference is one extra division by r! for combinations because we’re not counting the same group multiple ways.

How to tell them apart

The fastest test: ask “does rearranging count as different?” If yes → permutation. If no → combination.

• Lock combinations are actually permutations — 1-2-3 is different from 3-2-1.
• A committee of 3 students is a combination — the same 3 people are the same committee regardless of order.
• Finishing places in a race are permutations — 1st, 2nd, 3rd matters.
• A hand of cards is a combination — the same 5 cards are the same hand.

Worked example: permutation

How many ways can 5 runners finish in the top 3 positions?

Order matters (1st ≠ 2nd ≠ 3rd). Use P(5, 3) = 5! / (5 – 3)! = 120 / 2 = 60. So there are 60 possible 1st-2nd-3rd outcomes.

Worked example: combination

How many ways can you choose 3 students from a class of 5 for a committee?

Order doesn’t matter (any 3 chosen form the same committee). Use C(5, 3) = 5! / [3!·2!] = 120 / 12 = 10. So there are 10 possible committees.

Tricks for harder problems

1. Identify what you’re counting first, then ask “does order matter?”
2. For arrangements with restrictions (e.g., “two specific people must sit together”), treat the grouped people as one unit and multiply by the arrangements within the group.
3. For unordered selections from multiple groups (e.g., “2 boys and 3 girls from a class”), multiply the separate combinations.

Common mistakes

• Using permutation formulas for committee or team problems where order doesn’t matter.
• Forgetting the factorial in the denominator of the combination formula.
• Computing the factorial of n directly when you don’t need to — for P(20, 3) you only need 20 · 19 · 18 = 6,840, not the full 20!.