How to calculate $$\sum\limits_{k=0}^{mn} {mk1 \choose n1} {k+n \choose n}.$$

$\begingroup$ Mathematica finds $$\frac{\Gamma (m) \, _2F_1(n+1,nm;1m;1)}{\Gamma (n) \Gamma (mn+1)} $$ for this sum. $\endgroup$– user64494Apr 15 '19 at 2:57
An easy combinatorial way to see that $$\sum_{k=0}^{mn}\binom{mk1}{n1}\binom{k+n}n=\binom{m+n}{2n}:$$
The right hand side is the number of ways to pick a subset of size $2n$ from $\{1,2,\dots,m+n\}$.
The $k$ term on the left hand side is the number of ways to pick a subset of size $2n$ from $\{1,2,\dots,m+n\}$ whose $n$th element is $mk$.

3$\begingroup$ I cheated, of course, and reverse engineered the formula from T. Amdeberhan’s answer. $\endgroup$ Apr 14 '19 at 18:29

$\begingroup$ It's still a very nice combinatorial interpretation! @Jeremy Rickard $\endgroup$– luwApr 14 '19 at 19:59
This is the evaluation: $$\sum_{k=0}^{mn}\binom{mk1}{n1}\binom{k+n}n=\binom{m+n}{2n}.$$ There might be a direct connection to Vandermonde's identity but let's apply the socalled WilfZeilberger technique. To this end, divide through by the RHS and define the functions $$F(m,k)=\frac{\binom{mk1}{n1}\binom{k+n}n}{\binom{m+n}{2n}} \qquad \text{and} \qquad G(m,k)=\frac{F(m,k)\cdot(mn)!(mk)k}{(m+k+n1)(m+n+1)}.$$ Then, check routinely that (for instance, divide through by $F(m,k)$ on both sides) $$F(m+1,k)F(m,k)=G(m,k+1)G(m,k).$$ Now, sum both sides over all integers $k$ and notice that the RHS vanishes. That means $h(m+1)h(m)=0$ where $h(m)=\sum_kF(m,k)$. It remains to check that $h(n)=1$ which implies $h(m)=1$ for all $m$. This completes the proof.

$\begingroup$ Thank you! Could you give me some hint how to simplify it? I guess there might be some relation with Vandermonde's Identity. But I cannot connect them clearly. @T. Amdeberhan $\endgroup$– luwApr 14 '19 at 17:45
Yes, it is Chu  Vandermonde identity. We have ${mk1\choose n1}={mk1\choose mnk}=(1)^{mnk}{n\choose mnk}$, ${k+n\choose n}={k+n\choose k}=(1)^k{n1\choose k}$. Thus your sum equals $$ (1)^{mn}\sum_{k=0}^{mn} {n\choose mnk}{n1\choose k}= (1)^{mn}{2n1\choose mn}={m+n\choose mn}. $$