Permutations Calculator (nPr)

A permutation is an ORDERED arrangement of r items chosen from a pool of n. P(10, 3) = 720 means there are 720 distinct ways to award gold, silver, and bronze among 10 runners — gold-silver-bronze is different from bronze-silver-gold. Order matters.

A permutation cares about order; a combination doesn't. The same 3 students chosen from a class of 20: as a study group (combination) it's C(20, 3) = 1,140 distinct groups. As 3 distinct roles — president / VP / secretary (permutation) — it's P(20, 3) = 6,840 distinct assignments. Six times more, because every group of 3 can fill the roles in 3! = 6 different orders. Math relationship: P(n, r) = C(n, r) × r!.

P(n, n) is the full permutation — arranging ALL n items in order. For 5 books on a shelf, P(5, 5) = 5 × 4 × 3 × 2 × 1 = 120 = 5!. The formula confirms this: P(n, n) = n! / (n-n)! = n! / 0! = n! / 1 = n!.

P(n, 0) = 1 for any n. There is exactly one way to arrange zero items — the empty arrangement. The formula holds: P(n, 0) = n! / n! = 1.

Falling-factorial form. P(n, r) = n × (n-1) × ... × (n-r+1) — only r terms in the multiplication. P(52, 3) = 52 × 51 × 50 = 132,600. We never compute 52! directly. The downside: the result itself can still be huge, so we display it in compact scientific notation when needed.

Whenever the order of selection matters: race medals (gold ≠ silver), assigning distinct roles, seating arrangements, PIN codes (1234 ≠ 4321), word/letter arrangements, scheduling order. If "first / second / third" or "different roles" appears in the problem, it's a permutation.

Not directly. This calculator is for permutations WITHOUT repetition (each item used at most once). Permutations WITH repetition (like 4-digit PINs where digits CAN repeat) is just n^r — for example, 10^4 = 10,000 PINs if repeats are allowed. Use a basic calculator for that case.

Multiply r consecutive integers starting at n and going down. P(8, 3) = 8 × 7 × 6 = 336. P(15, 4) = 15 × 14 × 13 × 12 = 32,760. Don't plug into the n! / (n−r)! formula and compute the factorials separately — you'll just be cancelling most of it. The product form IS the cancellation.

Once the count exceeds 2^53 ≈ 9 × 10^15, JavaScript can't represent it exactly. The calculator flags it as approximate, but the magnitude (number of digits, leading digits, scientific-notation form) is reliable. Stirling's approximation in the right-hand panel gives an independent closed-form check.

Stirling's formula gives a closed-form estimate: n! ≈ √(2πn) × (n/e)^n. Applied to P(n, r) = n!/(n−r)! you get an estimate that needs no big-integer arithmetic. The relative error shrinks as ~1/(12n), so it's already under 1% by n = 9 and under 0.1% by n = 84. We surface it for huge inputs so you can sanity-check the magnitude.

P(52, 5) = 52 × 51 × 50 × 49 × 48 = 311,875,200. That's the number of distinct ordered 5-card draws from a deck. Most card-game problems use the unordered count C(52, 5) = 2,598,960 (= P(52, 5) / 5!).

Cryptography (key arrangements, password permutation counts), scheduling (ordered task slots), genetics (chromosome arrangements), bioinformatics (sequence alignment), routing (vehicle / TSP order), tournament brackets, lottery rankings. Any time an "order" attribute is meaningful, P(n, r) is the count.

For a multiset with n items where one element appears n₁ times, another n₂ times, etc., the count is n! / (n₁! × n₂! × …). Example: anagrams of MISSISSIPPI = 11! / (4! × 4! × 2!) = 34,650. This calculator handles all-distinct permutations only — divide by the repeat factorials yourself.