<s>[INST] Formula for partial sum of binoms: $\sum_{k=0}^{\lfloor n/2 \rfloor} \binom nk \binom mk$ I am dealing with some expressions containing combinatoric numbers. Does anybody know a formula for this?
$$\displaystyle\sum_{k=0}^{\left\lfloor \dfrac{n}{2} \right\rfloor} \binom{n}{k}\binom{m}{k}$$
 [/INST] Not simple and we need to consider three cases since we do not know the parity of $n$ and what $\left[ \dfrac{n}{2} \right]$ means (floor or ceiling function).


*

*Suppose that 
$n$ is odd $(n=2p+1)$ and that $\left[ \dfrac{n}{2} \right]=\left\lfloor \frac{n}{2}\right\rfloor=p$, the expression is
$$S_p=\sum_{k=0}^{p} \binom{2p+1}{k}\binom{m}{k}$$ and the result is given by $$S_p=\frac{\Gamma (m+2 p+2)}{\Gamma (m+1) \Gamma (2 (p+1))}-\binom{2 p+1}{p+1}
   \binom{m}{p+1} \, _3F_2(1,-p,-m+p+1;p+2,p+2;1)$$

*Suppose that $n$ is odd $(n=2p+1)$ and that $\left[ \dfrac{n}{2} \right]=\left\lceil \frac{n}{2}\right\rceil=p+1$, the expression is
$$S_p=\sum_{k=0}^{p+1} \binom{2p+1}{k}\binom{m}{k}$$ and the result is given by
$$S_p=\frac{\Gamma (m+2 p+2)}{\Gamma (m+1) \Gamma (2 (p+1))}-\binom{2 p+1}{p+2}
   \binom{m}{p+2} \, _3F_2(1,1-p,-m+p+2;p+3,p+3;1)$$

*If $n$ is even $(n=2p)$, whatever $\left[ \dfrac{n}{2} \right]$ could mean, the expression is
$$S_p=\sum_{k=0}^{p} \binom{2p}{k}\binom{m}{k}$$ and the result is given by 
$$S_p=\frac{\Gamma (m+2 p+1)}{\Gamma (m+1) \Gamma (2 p+1)}-\binom{2 p}{p+1} \binom{m}{p+1}
   \, _3F_2(1,1-p,-m+p+1;p+2,p+2;1)$$ In all cases, appears the generalized hypergeometric function.

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