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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?

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    This recent math.SE question asks for a proof of one of the formulas for the number of derangements: http://math.stackexchange.com/questions/14477/. – Mike Spivey Dec 17 '10 at 15:52
  • Check out http://en.wikipedia.org/wiki/Derangement – Florian Dec 17 '10 at 13:16
  • 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. – Holdsworth88 Jan 18 '12 at 01:41

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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.

Martin Sleziak
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Quixotic
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