Prove: ${n\choose 0}{m\choose k}+{n\choose 1}{m\choose {k-1}}+...+{n\choose k}{m\choose 0}={{m+n}\choose k}$
Is it possible to use Pascal's identity or writing binomial coefficients with factorials?
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1https://en.wikipedia.org/wiki/Vandermonde's_identity – Jean-Claude Arbaut Nov 02 '15 at 18:14
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1There is a combinatorial argument: taking $k$ items from $2$ piles respectively of $n$ and $m$ items is equivalent to take $i$ items in the first $n$-pile and $k-i$-items in the second $m$-pile. Then, you sum over all the possibilities for $i$. – MoebiusCorzer Nov 02 '15 at 18:15
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You can equate coefficients in $(1+x)^m(1+x)^n=(1+x)^{m+n}$, which is an algebraic way of expressing the combinatorial argument given by @MoebiusCorzer – Mark Bennet Nov 02 '15 at 18:17
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4This is a duplicate of Vandermonde's Identity : Summations with binomial coefficients and of this question, which has a nice combinatorial answer. – Brian M. Scott Nov 02 '15 at 20:11