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I know that the number of permutations with no fixed points over a set with $n$ elements approaches $\frac{n!}e$ as $n$ grows.

I'm interested in finding a limit (if there's exist) for the number of permutations with a single fixed point.

Thank you

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I think you mean "the number of permutations over a set with $n$ elements with no fixed points approaches $\frac{n!}{e}$." –  Qiaochu Yuan Jan 21 '12 at 18:57
    
corrected it... thanks. –  Amihai Zivan Jan 21 '12 at 19:04

1 Answer 1

up vote 6 down vote accepted

If $D(n)$ denotes the number of derangements of $n$ (permutations with no fixed points), then a permutation with a single fixed point is just a fixed point together with a derangement of the non-fixed points, so there are $n D(n-1)$ such permutations on $n$ letters. Hence there are asymptotically $\frac{n!}{e}$ such permutations as well.

A more general approach to questions of this type that also allows you to place constraints on the number of $k$-cycles for any $k$ is given by the exponential formula; see this series of blog posts for an introduction.

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Why D(n) is asymptotically n!/e? - Thanks –  Amihai Zivan Jan 21 '12 at 19:18
    
@Amihai: see for example en.wikipedia.org/wiki/Derangement . –  Qiaochu Yuan Jan 21 '12 at 19:22
    
Sorry, was not thinking. it's easy... thanks again! –  Amihai Zivan Jan 21 '12 at 19:28

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