# Number of permutations of $n$ where no number $i$ is in position $i$

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?

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.

• Please add more content from the cited page (e.g. describe derangements, etc). Otherwise, this should really be left as a comment. – robjohn Aug 7 '14 at 0:07

In his combinatorics lecture notes, Jacob Lurie analyses the problem of derangements. It is at the end of these note: http://www.math.harvard.edu/~lurie/155notes/lecture3.pdf

and continues into these