Counting the number of permutations of a set of n elements with k fixed points

Asked Feb 15 '17 at 22:52

Active Feb 21 '17 at 01:38

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I need to find the probability that a random permutation of n elements has exactly k fixed points but I am confused on how to do so.

I have found this formula but I don't understand how to derive it. $D_{n,k}= \frac{n!}{k!}\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$

I know there are $nCk$ choices for the k fixed points but don't know how to calculate the number of derangement's on the remaining $n-k$ elements. I know that the total number of permutations is $n!$ so I will need to divide the above formula by $n!$ to find the total probability.

edited Feb 21 '17 at 01:38

asked Feb 15 '17 at 22:52

Koala

The numerators (over $n!$) are usually called rencontres numbers or counts of partial derangements. Their derivation has been previously discussed here. – hardmath Feb 16 '17 at 01:54
This isn't really my thing, but you might consider looking at this recent paper, which is very pertinent to your question. – user397701 Feb 16 '17 at 02:28

Counting the number of permutations of a set of n elements with k fixed points

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