Your result will appear here
Fill in the fields and press Calculate.
A permutation counts arrangements where "order matters": choosing r items from n and lining them up, the order you place them in makes a different outcome — (A, B) and (B, A) are two distinct arrangements. From race finishes to password possibilities, from titled appointments to letter arrangements, the answer to "how many different orderings" is a permutation.
Enter the number of items (n) and how many to arrange (r), and P(n, r) is calculated instantly.
How is it calculated?
Formula
P(n, r) = n! ÷ (n−r)!. You place r items in order: n choices for the first spot, n−1 for the second, … n−r+1 for the rth — the product of those.
Permutation or combination?
Again the deciding question: does order matter? - If it does → permutation (first-second-third, a password, a seating order). - If it does not → combination (a team, group, selection).
Because orderings are counted, P(n,r) is always greater than or equal to C(n,r) — exactly r! times as large.
Special cases
- P(n, n) = n! — arranging all items is the factorial itself. Five people arrange in P(5,5) = 5! = 120 ways.
- P(n, 1) = n — there are n ways to pick and place a single item.
- P(n, 0) = 1.
Where it is used
- Ordering/ranking: how many ways the top three can be decided in a race.
- Passwords/codes: the number of non-repeating digit codes.
- Arrangements: books on a shelf, people in a line, letter permutations.
For unordered selections use a combination calculator; for the factorial underneath use a factorial calculator.
Worked example
In a race of 10 runners, how many ways can the top 3 (first, second, third) be filled? Order matters — who is first versus second changes the outcome — so it's a permutation: P(10, 3) = 10! ÷ 7! = 10 × 9 × 8 = 720. If instead you chose an unordered group of 3 from those 10, the answer would be C(10,3) = 120; the permutation being exactly 6 times larger (3! = 6) reflects that each group has 6 orderings. Arranging all 5 of a set gives P(5,5) = 5! = 120.
FAQ
How is a permutation calculated?+
With P(n, r) = n! ÷ (n−r)! — the number of ways to arrange r items from n in order. Just enter n and r.
What is the difference between a permutation and a combination?+
In a permutation order matters (A,B ≠ B,A); in a combination it does not. Since P(n,r) = C(n,r) × r!, a permutation is always greater than or equal to a combination.
Why does P(n, n) equal n!?+
Arranging all n items means n choices for the first place, n−1 for the second, and so on — the product is n!. For example all arrangements of 5 people are P(5,5) = 5! = 120.
Are password possibilities found with permutations?+
For passwords with no repeated digits, yes: a 4-digit code from 10 distinct digits is P(10,4) = 5,040. If digit repetition is allowed the count is 10⁴ instead (not a permutation).
Why is a race finish a permutation?+
Who takes first, second and third matters — order changes the outcome. So the top 3 from 10 runners can be filled in P(10,3) = 720 different ways.
Which tool do I use if order does not matter?+
If only who is selected matters (a group, team, committee), you need a combination. Use a combination calculator to find C(n, r).