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I've been working through Richard Hammack's book of proof, where I came across this problem relating to combinatorial proof.

show that:$$\sum_{k=0}^m {m \choose k}{n \choose p+k}={m+n \choose m+p}$$ I've managed to reduce the right hand side to $\sum_{k=0}^m{m\choose k}{n\choose m+p-k}$ by creating a set $S$ with $m+n$ elements, and breaking that into two subsets $A$ and $B$, and counting them separately, then using the Multiplication Principle to combine the results. I presume my problem lays in changing the $m+p-k$ into $p+k$, but any help is appreciated

James
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    The number of $k$-subsets of a set with $m$-elements is the same as the number $(m-k)$-subsets of the same set. The sum you want will be the sum you get when you replace $\binom{m}{k}$ with $\binom{m}{m-k}$ and reverse the order of summation. – shoteyes Jun 24 '21 at 02:13
  • @shoteyes thanks! That clears it up – James Jun 24 '21 at 16:27

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