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Number of permutations where n ≠ position n

There are $N!$ permutations of the set $\{1,2,\ldots,N\}$

How many of them have zero identity elements?

An identity element is an element that has a value equal to its position. ie When for some $i$, the ith element equals $i$.

For example, $(2,3,4,1)$ has no identity elements, whereas $(2,1,3,4)$ has two identity elements.

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marked as duplicate by Austin Mohr, Henning Makholm, hardmath, Hans Lundmark, Zev Chonoles Jun 16 '12 at 18:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

These are called derangements. This math.SE question, and this one, cover the answer to your question. – Zev Chonoles Jun 16 '12 at 16:08
Elements mapped to themselves by a permutation are called fixed points, not "identity elements". – hardmath Jun 16 '12 at 17:02
up vote 2 down vote accepted

The usual terminology would be "fixed points" or "1-cycles" instead of "identity elements".

A permutation with no fixed points is called a derangement. The Wikipedia article on derangements gives quite a bit of information about counting them.

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