Permutations and combinations calculator

How it works

Most counting questions come down to two choices: does the order matter, and may an object be repeated? Choosing $k$ things from $n$ gives four answers. With order and without repetition there are $P(n, k) = n (n-1) \cdots (n-k+1)$ arrangements; with order and repetition there are $n^k$ sequences; without order and without repetition there are $\binom{n}{k}$ combinations; without order but with repetition there are $\binom{n+k-1}{k}$ multisets, by the stars and bars argument.

Arranging all $n$ objects gives $n!$ permutations. When some objects are alike, orders that only swap alike objects are the same, which gives the multinomial coefficient $\frac{n!}{k_1! \, k_2! \cdots}$: the word MISSISSIPPI has $\frac{11!}{1! \, 4! \, 4! \, 2!} = 34650$ distinct anagrams. Around a round table, rotations are the same seating, which leaves $(n-1)!$. A derangement is an order where nothing is in its original place; there are about $n!/e$ of them.

In cryptography these numbers measure key spaces: a simple substitution cipher has $26! \approx 2^{88}$ keys, far too many to try one by one, which is why it is broken by frequency analysis instead.