Permutation Calculator P(n, r)
Count arrangements where order matters: r of n items in order, with repetition (passwords, PINs), words with repeated letters, and seats round a table (circular). Results are exact.
- Unordered choices C(n, r) = P ÷ r!10
- All n in order, n!120
- Number of digits2
How it is calculated
Permutations count arrangements where order matters: ABC and CBA are different. When order does not matter, use combinations, with P(n, r) = r! × C(n, r). Questions about words with repeated letters and seats round a table are exam favourites; the tool handles both directly.
P(n, r) = n! ÷ (n − r)! · With repetition: nʳ · Word: n! ÷ (k₁! × k₂! × …) · Circular: (n − 1)!n: number of items · r: places to fill · k₁, k₂, …: how many times each repeated character occurs.- P(n, r): n choices for the first place, n − 1 for the second, …, n − r + 1 for place r; multiply them.
- With repetition every place still has n choices: n × n × … = nʳ.
- For a word, the n! orders of its letters count each arrangement once for every swap of equal letters (k!), so n! is divided by those k!.
- Round a table rotations count as one, so one person is fixed and the other n − 1 are arranged.
Example
The first three places among 5 runners: P(5, 3) = 5 × 4 × 3 = 60. A 4-digit PIN: 10⁴ = 10,000. In ANKARA the letter A occurs three times: 6! ÷ 3! = 720 ÷ 6 = 120 arrangements. Six people can sit round a table in (6 − 1)! = 120 ways.
Frequently asked questions
How many arrangements does MATEMATİK have?
It has 9 letters, with M, A and T twice each: 9! ÷ (2! × 2! × 2!) = 362,880 ÷ 8 = 45,360. The words need not mean anything; every distinct order counts.
When are arrangements with repetition used?
When an item can fill more than one place: a 4-digit PIN from 0–9 has 10⁴ = 10,000 options, a 3-letter code from 26 letters 26³ = 17,576.
Why (n − 1)! for circular permutations?
Round a table, everyone moving one seat along leaves the same arrangement; the n rotations count as one, so n! ÷ n = (n − 1)!. For necklaces or key rings that can be flipped, mirror images count as one too, halving the result.
Why is P(n, 0) equal to 1?
There is one way to fill no places. For the same reason 0! = 1, and P(n, n) = n!.