Trying to work out a sum of the product of binomials that is not quite Vandermonde's identity, I know Vandermonde's identity gives:
$$
\binom{m+n}{r}=\sum_{k=0}^{r}{\binom{m}{k}\binom{n}{r-k}}
$$
I have something that is almost but not quite that, namely I have:
$$
\frac{\sum_{k=s}^{r}{\binom{k}{s}{\binom{m}{k}}{\binom{n}{r-k}}}}{\binom{m+n}{r}}
$$
For the special case where $s=0$, noting that $\binom{x}{0}=1,\space\forall{x}$, the sum collapses to Vandermonde's identity which cancels with the bottom giving $1$.
I've done some brute force evaluation up to 5 terms but I am not convinced of the pre-factor. I know I end up with something like:
$$
a_s\frac{\left(m\right)_i\left(n\right)_{r-s}}{\left(m+n\right)_s}
$$
But to be brutally honest I am some of those expansions are pages long and the chances of making a mistake or missing a term is ridiculously high. I think $a_s$'s are binomial coefficients but I am so non-confident of my rough work, I think it might be $\binom{r+1}{s}$ or $\binom{r}{s}$ or maybe $r\binom{r+1}{s}$. I am reasonably confident that there are running factorials and that the number of terms on the top matches the number of terms on the bottom.
Are there any identities I can use to:

*

*Figure out what the $a_s$ constants are

*Get to the running fractions from the sum


In seeking to evaluate
$${m+n\choose r}^{-1}
\sum_{k=s}^r {k\choose s} {m\choose k} {n\choose r-k}$$
we first note that
$${k\choose s} {m\choose k}
= \frac{m!}{s! \times (k-s)! \times (m-k)!}
= {m\choose s} {m-s\choose m-k}$$
so we get
$${m+n\choose r}^{-1} {m\choose s}
\sum_{k=s}^r {m-s\choose m-k} {n\choose r-k}.$$
The inner sum is
$$\sum_{k=0}^{r-s} {m-s\choose m-r+k} {n\choose k}
= \sum_{k=0}^{r-s} {m-s\choose r-s-k} {n\choose k}
\\ = [z^{r-s}] (1+z)^{m-s}
\sum_{k=0}^{r-s} z^k {n\choose k}.$$
Here the coefficient extractor enforces the upper limit of the sum and
we get
$$[z^{r-s}] (1+z)^{m-s}
\sum_{k\ge 0} z^k {n\choose k}
= [z^{r-s}] (1+z)^{m+n-s} = {m+n-s\choose r-s}.$$
We thus obtain the closed form
$${m+n\choose r}^{-1} {m\choose s}
{m+n-s\choose r-s}.$$
Here we presumably have $m+n\ge r$ or $m+n\lt 0$ to get a value we may
invert.
