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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.

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    The numerators (over $n!$) are usually called [rencontres numbers](https://en.wikipedia.org/wiki/Rencontres_numbers) or counts of partial derangements. Their derivation has been [previously discussed here](http://math.stackexchange.com/questions/17320/derivation-of-the-partial-derangement-rencontres-numbers-formula). – 2017-02-16
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    This isn't really my thing, but you might consider looking at [this][1] recent paper, which is very pertinent to your question. [1]: http://www.math.uiuc.edu/~ford/wwwpapers/fixed-point-permutation2.pdf – 2017-02-16

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