# Prove a combinatorial identity involving binomial coefficients: $\sum^n_{k=0}\binom{n}{k} \binom{m}{k}^{-1} = \frac{m+1}{m+1-n}$ for $m\geq n$

I encountered a problem as follows:

Show that $\sum^n_{k=0}\binom{n}{k} \binom{m}{k}^{-1} = \frac{m+1}{m+1-n}$ for $m\geq n$.

I wanted to use the binomial inversion formula, but I don't know where to start since the RHS is not a polynomial.

Any help would be appreciated.

HINT: Notice that

$$\frac{m+1}{m+1-n}\binom{m}n=\frac{m+1}{m-n+1}\binom{m}{m-n}=\binom{m+1}{m-n+1}\;,$$

something relatively nice, so let’s see if we can evaluate $\binom{m}n$ times the summation.

\begin{align*} \binom{m}n\sum_{k=0}^n\binom{n}k\binom{m}k^{-1}&=\sum_{k=0}^n\left(\frac{m!}{n!(m-n)!}\cdot\frac{n!}{k!(n-k)!}\cdot\frac{k!(m-k)!}{m!}\right)\\ &=\sum_{k=0}^n\frac{(m-k)!}{(m-n)!(n-k)!}\\ &=\sum_{k=0}^n\binom{m-k}{m-n}\\ &=\sum_{k=m-n}^m\binom{k}{m-n}\;. \end{align*}

(In each of the last two summations the upper number runs through the integers from $m-n$ to $m$ inclusive; it just does so in opposite orders.) Now use the hockey stick identity and a little algebra, and you’re home free.