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Goal: prove that $\displaystyle\sum_{i = 0}^{m}{\binom{k+i}{i} \binom{n-i}{m-i}} = \binom{n+k+1}{m}$

How to give a combinatorial argument, i.e. counting in two ways, for this problem? I tried by picking the least or largest element in the set, but it is just hard to get rid of the product of two binom on the left hand side. Any ideas would be appreciated.

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If we rewrite the identity as $$\sum_{i = 0}^{m}{\binom{k+i}{k} \binom{n-i}{n-m}} =\binom{n+k+1}{n+k+1-m}$$ then it can be proven by considering the value of the $(k+1)$-th element in the (sub)set of $n+k+1-m$ elements (which can only accept values from $k+1$ to $k+m+1$).

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