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 $\min\{2n,m\}$.



Sum of product of binomial coefficients
Mathematica tells me that


$$\sum\_{k=0, \, k =\text{odd}}^{2n} {2n \choose 2n-k}{2m-2n \choose m-k}=2 n \binom{2 m-2 n}{m-1}+\binom{2 m-2 n}{m}$$
$$\qquad+ \, \_4F\_3\left(\frac{1}{2}-\frac{m}{2},1-\frac{m}{2},\frac{1}{2}-n,1-n;\frac{3}{2},\frac{m}{2}-n+1,\frac{m}{2}-n+\frac{3}{2};1\right).\qquad(1)$$


Moreover,
$$\sum\_{k=0, \, k =\text{odd}}^{m} {2n \choose 2n-k}{2m-2n \choose m-k}=\binom{2 m-2 n}{m}$$
$$\qquad+\frac{2^{2 m-1}}{m!} \left(\frac{\Gamma \left(m+\frac{1}{2}\right)}{\sqrt{\pi }}-\frac{\Gamma \left(m-n+\frac{1}{2}\right)}{\Gamma \left(\frac{1}{2}-n\right)}\right), \qquad(2)$$
which differs from the answer $\frac 12 {2m \choose m}+(-1)^{m+1}2^{2m-1}{n-\frac 12 \choose m}$ given in the OP.  

For example, for $n=1,m=2$, both left-hand-side and right-hand-side of equation (2) evaluate to 5, while the answer in the OP evaluates to 4.


