Intuition for variant of Vandermonde identity

Question

I've been reading A Course in Enumeration by Martin Aigner, and have hit a stumbling block with understanding one of the examples. It begins:

Consider the following variant of the Vandermonde identity: $$\sum_{k=0}^{n}\begin{pmatrix}s+k \\k\end{pmatrix}\begin{pmatrix}n-k \\m\end{pmatrix}=\begin{pmatrix}s+n+1\\s+m+1\end{pmatrix} (s,m,n \in \mathbb{N}_0)$$

For starters, $(s+k)+(n-k)\ne s+n+1$, and from there on, my understanding crumbles.

Can this be explained for somebody who's just learned about the Vandermonde identity and who is relatively new to combinatorics in general? I managed to form an intuition about the original identity through the "Combinatorial proof" section of the Wikipedia page, for reference.

Answer 1

Consider binary words of length $s+n+1$ with $s+m+1$ $1$'s; these are clearly enumerated by $ \binom{s+n+1}{s+m+1}$.

Now consider the position of the $(s+1)^{th}$ $1$.

It will be preceded by $s$ $1$'s and $k$ $0$'s ($ \binom{s+k}{k}$).

It will be postceded by $m$ $1$'s and this chunk will be of length $n-k$ ($ \binom{n-k}{m}$).

Now $k$ can take any value from $0$ upto $n$, so \begin{eqnarray*} \sum_{k=0}^{n} \binom{s+k}{k}\binom{n-k}{m}=\binom{s+n+1}{s+m+1}. \end{eqnarray*}