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?
$$\displaystyle\sum_{k=0}^{\left\lfloor \dfrac{n}{2} \right\rfloor} \binom{n}{k}\binom{m}{k}$$

We can     find  a  closed form expression   in    case $m\leq  \left\lfloor\frac{n}{2}\right\rfloor$. We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n
\end{align*}

We obtain for $m\leq  \left\lfloor\frac{n}{2}\right\rfloor$:
  \begin{align*}
\color{blue}{\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n}{k}\binom{m}{k}}
&=\sum_{k=0}^m\binom{m}{k}\binom{n}{k}\\
&=\sum_{k=0}^m\binom{m}{k}[z^k](1+z)^n\\
&=[z^0](1+z)^n\sum_{k=0}^m\binom{m}{k}z^{-k}\\
&=[z^0](1+z)^n\left(1+\frac{1}{z}\right)^m\\
&=[z^m](1+z)^{m+n}\\
&\,\,\color{blue}{=\binom{m+n}{m}}
\end{align*}

