Proof of derangements identities

Question

Please help me proving this derangement identity. The problem is show that

$D(n,r,k) = rD(n-1,r-1,k) + D(n-1,r,k)$, where $r < n$*

Answer 1

I’m going to assume that as in D. Hanson, K. Seyffarth, and J.T. Weston, ‘Matchings, Derangements, Rencontres’, Mathematics Magazine, Vol. $56$, No. $4$, pp. $224$-$229$, $D(n,r,k)$ is the number of $r$-permutations of $[n]=\{1,\ldots,n\}$ with exactly $k$ fixed points, where an $r$ permutation of $[n]$ is a permutation $a_1a_2\ldots a_r$ of $r$ elements of $[n]$, and $i$ is a fixed point of $a_1a_2\ldots a_r$ iff $a_i=i$.

It’s helpful to notice that if $\sigma$ is any permutation of $[n]$, we could define $i\in[r]$ to be a fixed point of $a_1a_2\ldots a_r$ if $a_i=\sigma(i)$ without changing the value of $D(n,r,k)$. Imagine an exam with a list of $r$ terms and $n\ge r$ brief definitions that has exactly one correct definition for each term and possibly some extra definitions not matching any of the terms; $D(n,r,k)$ is the number of ways to match $k$ terms with their correct definitions and $r-k$ terms with incorrect definitions. The original definition in terms of $r$-permutations is based on arranging the $n$ definitions so that the $i$-th definition matches the $i$-the term for $i=1,\ldots,r$, but it clearly doesn’t matter where in the list the correct definitions actually are.

First notice that every $r$-permutation of $[n-1]$ with $k$ fixed points is also an $r$-permutation of $[n]$ with $k$ fixed points, and that every $r$-permutation of $[n]$ with $k$ fixed points that does not include $n$ is obtained in this way; this accounts for $D(n-1,r,k)$ $r$-permutations of $[n]$ with $k$ fixed points.

Now consider building an $r$-permutations of $[n]$ with $k$ fixed points that do include $n$. We can put the $n$ in any of the $r$ positions in the permutation. Can you finish this case? I’ve done so in the spoiler-protected area below, but you should try to finish it yourself before looking.

Since $n>r$, $n$ cannot be a fixed point of the permutation, so exactly $k$ of the remaining $r-1$ positions must be fixed points. Choosing which $k$ of those positions are fixed points determines those $k$ elements of the permutation, and the remaining $r-1-k$ elements must be chosen from $[n-1]$ in such a way that they are not fixed points; this can be done in $D(n-1,r-1,k)$ ways: it’s equivalent to the process of choosing an $(r-1)$-permutation of $[n-1]$ with $k$ fixed points. This accounts for the other $rD(n-1,r-1,k)$ $r$-permutations of $[n]$ with $k$ fixed points.