$W_n^{(k)}$ is the number of permutations in a set of all $n!$ permutations in a $n$-element set which has $k$ fixed points.
$W_n^{(0)}$ is the number of n-derangements where $\frac{W_n}{n!} \rightarrow \frac{1}{e}$ as $n \rightarrow \infty$
I need to show that $$W_n^{(k)}={{n}\choose{k}} \cdot W_{n-k}$$
From this formula, I should be able to show, the probability that a random permutation in a set of all n! permutations in an $n$-element set which has $k$ fixed points converges to $\frac{1}{e \cdot k!}$
This is my attempt so far:
Since $W_n^{(k)}$ is just # of permutation of n elements with k fixed points, we can say $W_n^{(k)} = \frac{n!}{(n-k)!}$. Then the number of distinct subsets with k elements that can be chosen from a set with n elements is $\frac{n!}{k!(n-k)!} \equiv$ ${{n}\choose{k}}$
So now, I attempt to try multiplying $\frac{n!}{k!(n-k)!} \cdot W_{n-k}$ and try to find the limit of it. But what is $W_{n-k}$? Any ideas, tips ,help would be appreciated! Thanks.
The number of permutations with exactly $k$ fixed points is $\binom{n}{k}$ times the number of derangements of $n-k$ objects. You know that for fixed $k$ and large $n$ the number of derangements of $n-k$ objects behaves like $(n-k)!\cdot \frac{1}{e}$.
The above paragraph can be summarized in symbols as follows: $$\frac{W_n^{(k)}}{n!}=\frac{1}{n!}\binom{n}{k}(n-k)!\frac{W_{n-k}^{(0)}}{(n-k)!}.$$
The product of the first three terms on the right simplifies to $\frac{1}{k!}$, and the last term has limit $\frac{1}{e}$.