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Proof that:

$$\sum_{k=0}^m \binom{m}{k}\binom{n+k}{m} = \sum_{k=0}^m \binom{n}{k}\binom{m}{m-k}2^k$$ for $k,m,n \in \mathbb{N}$.

Can someone help me understand this equality by giving a combinatoric argument?

I believe that the RHS means "all possible ways to first select $k$ elements from $m$ elements and thereafter, selecting $m$ elements out of $n+k$ elements". Same for the LHS. The factor $2^k$ probably has to do something with having to pick between two possible options for $k$ elements.

I can't really find a good scenario to prove the equation. I tried with a group of $m$ kids, with $n$ boys and $k$ girls. But that did not really work out.

Thanks for helping!

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  • $\begingroup$ What have you tried? $\endgroup$ – Phicar Jan 22 at 2:52
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Well suppose you have $m$ boys and $n$ girls and you want to pick a team of $m$ people. Assume, further, that the way to select them is as follows, you first pick $k$ man in a preliminary round $\binom{m}{k}$ and in a secondary round out of the chosen man and all the woman you pick your team of $m$ i.e., $\binom{n+k}{m}.$ Now, assume you pick directly the man and the women of the team, so they have to add up to $m$ so $\binom{n}{k}\binom{m}{m-k}$ but you have to pay the men who pass the preliminary round in the LHS, and so you can choose any subset of the man who were not selected ($k$ of them) in $2^k$ ways.

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