22
$\begingroup$

I am trying to figure out how many permutations exist in a set where none of the numbers equal their own position in the set; for example, $3,1,5,2,4$ is an acceptable permutation where $3,1,2,4,5$ is not because 5 is in position 5. I know that the number of total permutations is $n!$.
Is there a formula for how many are acceptable given the case that no position holds its own number?

$\endgroup$
23
$\begingroup$

What you are looking for is known as derangement. However, for counting the number of derangement for say $n$-elements you could possible use a trick, compute$\frac{n!}{e}$ and then round off to an integer and this will give you the desired result.

This is actually another application of $e$, which was discovered by Jacob Bernoulli in the problem of derangement, also known as the hat check problem.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy