How this instrument works
A permutation is an ordered arrangement: pick r items out of a pool of n, and count arranging them in a different order as a different outcome. Filling r open slots one at a time drains the pool as you go — the first slot has n choices, the second has n − 1 since one item is already placed, the third has n − 2, and so on until r slots are filled. Multiplying that shrinking run of choices together gives nPr, and the same running product is written compactly as n! divided by (n − r)!, since dividing out the untouched tail of the factorial leaves exactly that chain of n, n − 1, n − 2, and so on down to n − r + 1.
The defining test for whether a problem calls for a permutation is simple: swap two of the selected items and ask whether that counts as a new outcome. Gold, silver and bronze among a field of racers clearly does — the same three racers finishing in a different order is a different result — so that is a permutation problem. Picking three racers to send to a different meet, with no medal distinction, is not — the group is the same regardless of the order they were chosen in, which is what the related combination calculation is for.
nPr grows fast. Each extra racer or item multiplies the count by roughly n again, so pools that feel modest in size — a dozen finalists, a shelf of twenty books — already produce arrangement counts in the millions once r gets past a handful. That growth is exactly what the factorial in the formula captures: n! counts every possible full ordering of the whole pool, and nPr is the slice of that count corresponding to filling only the first r positions.
- Enter the size of your full pool into n (total items) — everything available to be arranged, whether that's racers, applicants, or letters.
- Enter how many of them are actually being placed into ordered positions into r (items arranged).
- Read nPr — permutations directly beneath the inputs; the instrument runs n! / (n − r)! for you.
- If r is larger than n, the instrument flags it — you cannot arrange more items into positions than the pool actually contains.
- Set r equal to n to get every possible full ordering of the entire pool, which reduces the formula to n! itself.
Worked example — medal order among 10 racers
Ten racers cross the finish line, and three medals — gold, silver, bronze — are awarded in order. Enter n = 10 and r = 3. The instrument computes 10! divided by (10 − 3)!, which is 10! divided by 7!. Written as a shrinking chain of choices, that's 10 racers who could take gold, then 9 remaining who could take silver, then 8 remaining who could take bronze: 10 × 9 × 8 = 720. nPr reads 720.
That figure counts every distinct ordered medal outcome — a different set of three racers, or the same three in a different order, both count as separate results. If the question instead asked only which three racers make the podium, with no distinction between the medals, the count drops to 120 — the combination total, exactly 3! = 6 times smaller, since each group of three can be arranged across gold, silver and bronze in 6 different orders.
Questions
What does nPr actually count?
The number of ways to choose r items from a pool of n and arrange them in order, where a different order counts as a different outcome. For n = 10 and r = 3, nPr = 720 — every distinct ordered way three of the ten items could be picked and placed into first, second and third position.
What's the difference between a permutation and a combination?
A permutation counts order as significant — arranging, ranking, assigning distinct roles. A combination ignores order and only counts which items ended up selected. Awarding gold, silver and bronze among 10 racers is a permutation (720 outcomes); picking an unordered group of 3 finalists from the same 10 is a combination (120 outcomes) — 6 times smaller, since each group can be ordered 3! = 6 ways.
Why does the formula divide by (n − r)!?
Filling r slots one at a time uses up the pool item by item and then stops, leaving n − r items untouched. The full factorial n! counts every ordering of the whole pool, including all the ways to arrange that untouched leftover; dividing by (n − r)! cancels those leftover orderings back out, leaving only the count for the r slots that were actually filled.
What happens when r equals n?
The formula reduces to n! / 0!, and since 0! = 1 by definition, that's simply n! — every possible full ordering of the entire pool. Arranging all 5 of 5 items, for example, gives 5! = 120 orderings.
Can r be larger than n?
No — you cannot arrange more items into positions than the pool contains, since a permutation without repetition uses each item at most once. This instrument flags that case rather than returning a number.
How is this different from the permutation-with-repetition calculator?
This calculator assumes each item is used at most once — once a racer takes gold, they're no longer available for silver. If repetition is allowed instead — like a PIN code where the same digit can appear in multiple positions — the count follows n^r instead, which is what the permutation-with-repetition instrument computes.