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$${m+n \choose m+r} = \sum\limits_{k=0}^{m}{m \choose k}{n \choose r+k}$$ The problem makes sense to me, intutively, if I write out the left-hand side as: $${m\choose m}*{n \choose r}+{m\choose m-1}*{n \choose r+1}+...+{m\choose 1}*{n \choose r+m-1}+{m\choose 0}*{n \choose r+m}$$ You've introduced a group of size m to a group of size n and you have to pick m plus r in the first 'round', you simply pick every element from group m and get m elements, plus pick r elements from group n. Then the next time you don't pick one m and you pick an extra element from n, increasing in this way until you pick m+r elements from group n. The trouble is I have no idea how to write this out mathematically.

Please don't mark this as a duplicate of a post about Vandermonde's identity without explaining how this problem can be solved using Vandermonde's identity.

Arnaud D.
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    What do the double parentheses mean in the second factor on the right of the sum? – coffeemath Oct 10 '19 at 16:39
  • ...what you've just described sounds like a bijection proof to me. You could describe it in terms of colors of elements if you wanted to, but you've shown a way of counting a specific thing that yields the value on the left, and a different way of counting that thing that yields the value on the right. – Steven Stadnicki Oct 10 '19 at 16:39

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