Vandermonde's identity gives $\sum_{k=0}^r \binom{m}{k}\binom{n}{r-k}=\binom{m+n}{r}.$
Here is an example of Vandermonde's-like identity: For all $0 \le m \le n$, $\sum_{k=0}^{2m} \binom{\left\lfloor\frac{n+k}{2}\right\rfloor}{k}\binom{m+\left\lfloor\frac{n-k}{2}\right\rfloor}{2m-k}=\binom{m+n}{2m}$ (Note that $\left\lfloor\frac{n+k}{2}\right\rfloor+\left(m+\left\lfloor\frac{n-k}{2}\right\rfloor\right)$ is either $m+n$ or $m+n \pm 1$)
I wonder if there are some similar identities where $m(k)$ and $n(k)$ are functions of $k$ and $m(k)+n(k)$ is 'almost' constant, says $m+n$, the identity looks like
$\sum_{k=0}^r \binom{m(k)}{k}\binom{n(k)}{r-k}=\binom{m+n}{r}?$