Number of permutations where n ≠ position n

Question

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?

Answer 1

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

http://www.math.harvard.edu/~lurie/155notes/lecture4.pdf.

Answer 2

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.

Answer 3

Check out http://en.wikipedia.org/wiki/Derangement