How this instrument works
Both formulas answer the same underlying question — how many ways can a smaller group be picked from a larger pool? — and they split entirely on whether the order of picking counts as a different outcome. Filling the open slots one at a time uses up the pool bit by bit: the first slot has every item available, the second has one fewer to choose from, and so on down to the last slot, and multiplying those shrinking counts together gives the permutation total, the tally of every ordered selection. Each of those ordered lists can be reshuffled internally without changing which items got picked, so dividing that ordered tally by the count of internal reshuffles collapses the duplicates and leaves the combination total — the count of the distinct groups themselves, with order thrown away.
The two totals behave differently as the pick count sweeps from empty to everything, with the pool size held fixed. The combination total is symmetric and rises to a single peak near the halfway point before falling back to one, because keeping some items is the same act as setting the rest aside, and the two framings must always agree. The permutation total never falls: one more slot to fill can only add more ordered lists, so it climbs the whole way from a lone empty selection up to every possible full-pool ordering, flattening only on the very last step, since there is nothing left over to shuffle once just one or zero items remain untouched.
The common slip is reaching for the wrong one. Scoring a race's medal order with the unordered formula undercounts it, since that formula silently treats gold, silver, and bronze as interchangeable; counting an unordered committee with the ordered formula overcounts it, since every reshuffle of an identical group then gets counted as a separate committee. A quick check catches either mistake: whenever the pick count sits at zero or one, the two totals come out equal, since there is only a single way to arrange nothing or a lone item — the two formulas only pull apart once two or more items are being picked.
- Set n to how many distinct items sit in the pool overall — everything you could possibly draw from.
- Enter how many are being selected into r; the sheet flags it if r is set larger than n.
- Read Permutations, P(n,r) for the count when the pick order matters — arranging, ranking, assigning.
- Read Combinations, C(n,r) for the count when order is irrelevant — grouping, drawing, forming a subset.
- Compare the two outputs: dividing the permutations figure by the combinations figure always returns r! exactly.
Worked example — picking two runners from a squad of five
A relay coach has five runners trying out for two legs of a race, and the two legs are different jobs — lead-off and anchor — so which runner lands where matters. Any of the five runners could take the lead-off leg; whichever runner ends up there, four runners remain as candidates for the anchor leg, since that one is already spoken for. Multiplying those two counts together — five runners paired against four remaining candidates — gives twenty ordered assignments, and the permutation formula lands on the same figure directly: 5! divided by 3! is 120 divided by 6, which is 20.
Now the coach asks a second, different question: how many possible two-runner pairs could be sent to the meet at all, before any leg gets assigned? Order no longer matters, so the combination formula applies: 5! divided by the product of 2! and 3! is 120 divided by 12, which is 10. Ten distinct pairs are possible, and each of those ten pairs corresponds to exactly two of the twenty ordered assignments above — one for each way that pair could be split between lead-off and anchor, which is exactly why twenty divided by ten equals two, the value of 2!.
Questions
What is the difference between a permutation and a combination?
A permutation counts selections where order matters — arranging, ranking, or assigning distinct roles — while a combination counts the same selections with order ignored, just which items ended up chosen. For a pool of five with two picked, this sheet returns twenty ordered outcomes and ten unordered ones, the exact figures shown for that pair.
How exactly is P(n,r) related to C(n,r)?
P(n,r) = r! × C(n,r) for any valid pair of inputs. Every unordered group counted by the combination formula can be arranged internally in r! different orders, and each order is a separate outcome the permutation formula counts, so multiplying the group count by r! recovers the ordered count exactly — ten groups times two orderings gives the twenty permutations for a pool of five with two picked.
Why does the permutation formula divide by (n−r)!?
Filling the open slots one at a time uses up the pool item by item, stopping once every slot is filled. That running product equals the full factorial with an untouched tail removed — the factorial of whatever was left over and never picked — and dividing the full factorial by that leftover factorial is exactly how the unused tail gets cancelled back out.
What happens at the edges, when nothing or everything is picked?
Both formulas return one at the empty pick, since there is exactly one way to select and arrange nothing at all. Picking everything still leaves only one full-pool group, so the combination total stays one, but the permutation total equals the full factorial, since every ordering of the whole set counts as distinct. For a pool of five that means one hundred twenty orderings against a single group.
Is a permutation always a bigger number than a combination?
Only once two or more items are being picked. At zero or one pick the two formulas agree exactly, because there is nothing left to divide out. From two picks onward, the internal-reshuffle factor climbs past one and the ordered total pulls ahead of the unordered total by exactly that growing factor.
What's the most common mistake people make between the two?
Using the wrong one for the real situation being counted. Scoring a race's medal order with the unordered formula undercounts it, treating gold, silver, and bronze as interchangeable; counting an unordered committee with the ordered formula overcounts it, treating every reshuffle of an identical group as a separate committee. If the roles or ranks differ, count with order; if only membership matters, ignore it.