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 $$ \sum_{k=0, \, k =odd}^m {2n \choose 2n-k}{2m-2n \choose m-k}= \frac 12 \sum_{k=0}^m {2n \choose 2n-k}{2m-2n \choose m-k}-\frac 12 \sum_{k=0, }^m (-1)^k{2n \choose 2n-k}{2m-2n \choose m-k} $$ 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 $\min\{2n,m\}$.