The whole difference between a permutation and a combination fits in one word: order. A permutation is a list, where the sequence matters — first, second, third are all distinct roles. A combination is a group, where the sequence is irrelevant — three names tossed in a bag are the same three names no matter how you read them out. Get that straight and the formulas stop being something to memorise and start being something you can rebuild from scratch.
There’s a famous bit of evidence that people mix these up constantly: the “combination lock.” If 4-2-7 opens it and 7-2-4 doesn’t, then order clearly matters, which makes it a permutation lock. The name is wrong, and the mistake is the exact one this article fixes.
The one question that settles it
Before reaching for any formula, ask a single thing: if I swap two of my chosen items, do I get a different answer? If yes, order matters and you want a permutation. If swapping changes nothing, order doesn’t matter and you want a combination. Naming a President and a Vice-President is a permutation, because trading their titles produces a different outcome. Picking two people for a committee is a combination, because the committee is identical either way. Every problem in this area is really just that question in disguise.
Permutations: when order is everything
Picture eight contestants competing for gold, silver, and bronze. The medals rank the winners, so the order they’re handed out absolutely matters — this is permutation territory.
Count it by filling the podium one place at a time. For the gold there are eight choices. Once it’s awarded, seven people remain in the running for silver. With two medals gone, six are left for bronze. The total is 8 × 7 × 6 = 336 ways to fill the three places.
Now connect that to factorials, because it reveals where the formula comes from. The full factorial 8! multiplies 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, which goes too far — you only wanted the first three terms. The leftover tail, 5 × 4 × 3 × 2 × 1, is just 5!, and 5 is what remained after three medals were taken from eight. Dividing cancels the tail cleanly: 8! / 5! = 8 × 7 × 6 = 336. Since 5 is 8 − 3, the pattern generalises. To pick k items from n in a specific order:
P(n, k) = n! / (n − k)!
The formula isn’t a rule to trust on faith. It’s the “start with everything, stop when you run out of slots” idea written compactly.
Combinations: when order drops out
Change the prize. Instead of ranked medals, you’re handing three identical tin cans to eight people. Now the order you pick them in means nothing — giving cans to Alice, then Bob, then Charlie produces the same trio as Charlie, then Alice, then Bob. They’re equally unimpressed either way.
That sameness creates duplicates in the permutation count. Any group of three people can be arranged in 3 × 2 × 1 = 6 orders, and all six describe the same group. So the 336 ordered outcomes count every real group six times over. Divide the redundancy out: 336 / 6 = 56 actual combinations.
The general move is the same. Take the permutation count and divide by k!, the number of ways the chosen k items can be shuffled among themselves:
C(n, k) = P(n, k) / k! = n! / ((n − k)! · k!)
This quantity shows up so often it has its own symbol, the binomial coefficient, written as “n choose k.” Whenever you see it, read it as “the number of groups of size k you can pull from n, ignoring order.”
Why combinations always come out smaller
A useful sanity check follows directly from that division. For any set of items, there are always at least as many permutations as combinations, and usually far more, because every single combination expands into k! different permutations once you start caring about order. Three chosen people are one combination but six permutations. Choosing then ranking is strictly more work than just choosing, so the ordered count is always the larger of the two. If you ever calculate a combination and it comes out bigger than the matching permutation, you’ve made an error somewhere — the relationship can’t run that way.
The same numbers, side by side
The clearest way to feel the gap is to run identical situations through both lenses. Picking 3 people from 10 plays out very differently depending on whether their roles are interchangeable:
- A team of 3 from 10 — order irrelevant — is C(10, 3) = 120.
- A President, VP, and Waterboy from 10 — order defines the roles — is P(10, 3) = 720.
- Choosing 3 desserts from a menu of 10 to eat is C(10, 3) = 120.
- Listing your 3 favourite desserts in ranked order from that menu is P(10, 3) = 720.
Same people, same menu, same “3 from 10.” The only thing that changed was whether sequence carried meaning, and the answer jumped sixfold — exactly the 3! redundancy at work.
Making it stick
Skip the instinct to memorise two formulas as separate facts. They’re one idea: count the ordered arrangements, then divide away the orderings you don’t care about. Build the permutation by filling slots one at a time, and reach the combination by dividing by k!.
In practice, you’ll almost never need to recall the symbols cold. You’ll need the habit of pausing on a problem and asking whether swapping two picks changes the outcome. Ranked results, assigned roles, sequences, and yes, lock codes are permutations. Teams, hands of cards, committees, and shopping baskets are combinations. Decide that first, and the right formula — and the right answer — falls out on its own.