TITLE: Combinatorial proof involving factorials
QUESTION [7 upvotes]: Show that $$\sum_{k=0}^{m} \frac{m!(n-k)!}{n!(m-k)!} = \frac{n+1}{n-m+1}$$

I tried thinking of what the LHS summand could possibly be (permutation, combination), but I can't come up with anything; nor was I able to come up with a combinatorial proof.

REPLY [2 votes]: First rearrange the identity:
$$\sum_{k=0}^m\frac{(n-k)!}{(m-k)!}=\frac{(n+1)!}{m!(n+1-m)}=\binom{n+1}m(n-m)!\;.$$
Then note that
$$\frac{(n-k)!}{(m-k)!}=(n-k)^{\underline{n-m}}\;,$$
where $x^{\underline n}$ is the falling factorial, so that the identity can be written
$$\sum_{k=0}^m(n-k)^{\underline{n-m}}=\binom{n+1}m(n-m)!\;.\tag{1}$$
The term $(n-k)^{\underline{n-m}}$ is the number of ways to choose and permute $n-m$ elements of $[n-k]$, where $0\le k\le m$. Alternatively, it’s the number of ways to choose a set of $n-m+1$ elements of $[n+1]$ whose largest element is $n-k+1$ and permute the other $n-m$ elements of the set. Thus, the lefthand side of $(1)$ counts the ways to choose $n-m+1$ elements of $[n+1]$ and permute the $n-m$ smallest of the chosen elements.
On the other hand, there are $\dbinom{n+1}{n-m+1}=\dbinom{n+1}m$ ways to choose $n-m+1$ elements of $[n+1]$, and there are then $(n-m)!$ ways to permute the $m-n$ smallest of the chosen elements, so the righthand side of $(1)$ counts the same thing, and the identity is established.