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I would like to compute the following sum: $$ \sum_{k=0, \, k =odd}^{\min\{2n, m\}} {2n \choose 2n-k}{2m-2n \choose m-k} $$ So far I can prove that $$ \sum_{k=0, \, k =odd}^m {2n \choose 2n-k}{2m-2n \choose m-k}=\frac 12 {2m \choose m}+(-1)^{m+1}2^{2m-1}{n-\frac 12 \choose m}. $$ which can be proven by splitting sum as $$ \sum_{k=0, \, k =odd}^m {2n \choose 2n-k}{2m-2n \choose m-k}= \frac 12 \sum_{k=0, \, k =odd}^m {2n \choose 2n-k}{2m-2n \choose m-k}-\frac 12 \sum_{k=0, \, k =odd}^m (-1)^k{2n \choose 2n-k}{2m-2n \choose m-k} $$ and computing first sum using Chu-Vandermond identity and second -- using notion of coefficient-extractor.

I am not sure on how to proceed when the upper bound of summation is $\min\{2n,m\}$.

user124297
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  • How do you define ${a \choose b}$ when $a<0$? You are going to have $2m-2n<0$ in your second binomial factor when $m < n$. – VIVID Oct 15 '20 at 17:47
  • Yes, I am assuming that it is zero. Thank you – user124297 Oct 15 '20 at 17:49
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    Cross-posted: https://mathoverflow.net/questions/374163/sum-of-product-of-binomial-coefficients – RobPratt Oct 15 '20 at 18:24
  • This may be helpful: https://math.stackexchange.com/questions/80649/combinatorial-proof-that-sum-limits-k-0n-binom2kk-binom2n-2kn-k – VIVID Oct 15 '20 at 18:37
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    Some two months ago I posted a proof for the upper limit m, which was deleted. Also, when you use complex residues the case of the second argument of a binomial coefficient being negative yields zero, which in your case means no contribution when $k\gt m$ or $k\gt 2n$. The upper limit is enforced automatically. I am still in possession of the source of the deleted post. – Marko Riedel Oct 15 '20 at 21:43

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