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Let $n\in\Bbb N$, $k\in\{0,\dots,2n\}$. Is this equation valid? $$\binom{2n}{k}=\sum_{i=0,\dots,n\\ j=0,\dots,n\\i+j=k}\binom{n}{i}\binom{n}{j}$$

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closed as off-topic by Ahmad Bazzi, Jendrik Stelzner, user91500, Jyrki Lahtonen, Brian Borchers Sep 8 '18 at 19:28

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Yes, it holds. We have that $$\sum_{i=0,\dots,n\\ j=0,\dots,n\\i+j=k}\binom{n}{i}\binom{n}{j}=\sum_{i=0,\dots,n\\ j=0,\dots,n\\i+j=k}[x^i](1+x)^{n}\cdot [x^j](1+x)^{n}=[x^k](1+x)^{2n}=\binom{2n}{k}.$$

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  • $\begingroup$ Many thanks for such a beautiful explanation. $\endgroup$ – szw1710 Sep 21 '17 at 10:53

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