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$$\sum_k {m\choose k} {n \choose k} = {m+n \choose n}$$

In this identity we seem to be choosing subsets that do $\it not$ contain k of type m and type n for all possible k. In the style of Vandermonde's identity we are then choosing n elements of both types to not be in the set. Where might I go from here to form a full proof?

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A special case of… – Timere Nov 16 '13 at 10:26
up vote 5 down vote accepted

I guess that rewriting the identity as $$ \sum_{k} \binom{m}{k}\binom{n}{n-k} = \binom{m+n}{n} $$ it should be clear.

You are choosing $n$ elements in a set that is the disjoint union of a set $A$ of $m$ elements and a set $B$ of $n$ elements. So for each $k$ you choose $k$ elements from $A$, and the rest from $B$.

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Yes, it is quite obvious now thanks, I can't believe I missed that. – 114 Apr 22 '13 at 12:10
@Stopwatch, you're welcome. I believe very few people can fairly say they never miss the obvious. – Andreas Caranti Apr 22 '13 at 12:19

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