<s>[INST] Prove that $\sum_{0\leq k\leq r} \binom{n+k}{k} \binom{m+n+k}{n+k} = \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}$ where $m,n,r\in \mathbb{N} $ $\sum_{0\leq k\leq r} \binom{n+k}{k} \binom{m+n+k}{n+k} = \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}$ where  $m,n,r\in \mathbb{N} $. 
Exam problem which stayed unproven for me. I tried induction/binomial properties together and separately.I also tried to find a power serie which could use these binomials. Could someone provide at least a hint? 
 [/INST] Induction always works for hypergeometric sum identities.
$r=0$: $$\sum_{k=0} \binom{n+k}{k} \binom{m+n+k}{n+k} = \binom{m+n}{n} = \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}$$
$r > 0$: $$\begin{eqnarray}
\textrm{LHS} &=& \binom{m+n}{n} \binom{m+n+(r-1)+1}{m+n+1} + \binom{n+r}{r} \binom{m+n+r}{n+r} \\
&=& \binom{m+n}{n} \frac{(m+n+r)!}{(m+n+1)!(r-1)!} + \frac{(m+n+r)!}{n!r!m!} \\
&=& \binom{m+n}{n} \left[ \frac{(m+n+r)!}{(m+n+1)!(r-1)!} + \frac{(m+n+r)!}{(m+n)!r!} \right] \\
&=& \binom{m+n}{n} \left[ \frac{(m+n+r)!r + (m+n+r)!(m+n+1)}{(m+n+1)!r!} \right]\\
&=& \binom{m+n}{n} \left[ \frac{(m+n+r)!(r+m+n+1)}{(m+n+1)!r!} \right]\\
&=& \binom{m+n}{n} \left[ \frac{(m+n+r+1)!}{(m+n+1)!r!} \right]\\
&=& \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}
\end{eqnarray}$$

Alternatively, for a combinatorial proof, both sides count the number of ways of colouring $m+n+r$ objects in a line such that $m$ are red, $n$ are blue, the rest are green or yellow, and all the yellow ones are together at the end of the line.
On the LHS, $k$ is the number of greens and the term inside the sum is a simple multinomial.
On the RHS, we take $m+n+r+1$ slots and choose $m+n+1$ of them. Colour the slots to the right of the last chosen one yellow, and the unchosen slots to its left green. Then delete that slot. From the other $m+n$ chosen slots, choose $m$ to colour red.
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