The one question that decides which formula you need
A permutation counts arrangements where order matters — picking a 1st, 2nd and 3rd place winner from ten runners gives a different result for (Alice, Bob, Carol) than for (Carol, Bob, Alice), even though the same three people are involved, because who came first is a different outcome from who came third. A combination counts selections where order doesn't matter — picking a 3-person committee from ten candidates gives the exact same result whether you name Alice first or Carol first, because a committee of {Alice, Bob, Carol} is the same committee no matter what order you listed the names in.
That's the entire distinction. Every other difference between the two formulas is just the mathematical consequence of that one question, not a separate rule to memorise.
The formulas, and why one has an extra term
| Permutations | Combinations | |
|---|---|---|
| Does order matter? | Yes | No |
| Formula (choosing r from n) | n! / (n - r)! | n! / (r! × (n - r)!) |
| Typical example | Awarding 1st, 2nd, 3rd place | Choosing a committee, a hand of cards, a set of toppings |
| Relationship | Always ≥ the matching combination count | Permutations ÷ r! (the ways to reorder the r chosen items) |
The combination formula is just the permutation formula divided by r! (r factorial) — the number of ways to reorder whatever r items you've already chosen. That division is doing exactly one job: collapsing every reordering of the same chosen group down into a single count, because a combination doesn't care about order in the first place. If you understand permutations, you already understand combinations — they're the same count with all the duplicate reorderings removed.
A worked example that shows both at once
Take 5 people and ask two different questions about groups of 3. First: in how many different orders can 3 of them finish a race (1st, 2nd, 3rd)? That's a permutation, because finishing 1st is a different outcome from finishing 3rd even with the same three runners: 5!/(5-3)! = 5×4×3 = 60 possible orderings.
Second: in how many different ways can you choose a 3-person team from the same 5 people, with no roles or ranking? That's a combination, because a team of {Alice, Bob, Carol} is one single outcome regardless of who you list first: 5!/(3!×2!) = 10 possible teams. Notice 60 divided by 3! (which is 6, the number of ways to reorder 3 people) gives exactly 10 — the permutation count collapsed down by removing every reordering of the same three people.
The everyday test that avoids the mix-up
Before reaching for either formula, ask: if I swapped the order of the items I picked, would that count as a genuinely different result, or exactly the same one? A lottery draw of 6 numbers is a combination, because {3, 17, 22, 41, 45, 49} is the same winning set no matter what order the balls came out in. A 4-digit PIN code is a permutation-style count (more precisely, arrangements with repetition allowed), because 1234 and 4321 are two completely different PINs even though they use the same four digits.
Both formulas above assume you're choosing without repetition - each item can only be picked once. Situations that allow repeats (like PIN digits, where the same digit can appear more than once) use a related but different formula.
Where this comes from
- The formulas for permutations (nPr) and combinations (nCr) are standard combinatorics, taught identically across essentially every introductory probability and statistics curriculum.