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?
Number of permutations of $n$ where no number $i$ is in position $i$
20
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combinatorics
permutations
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1This recent math.SE question asks for a proof of one of the formulas for the number of derangements: http://math.stackexchange.com/questions/14477/. – 2010-12-17