TITLE: Half of Vandermonde's Identity
QUESTION [5 upvotes]: We know Vandermonde's Identity, which states 
$\sum_{k=0}^{r}{m \choose k}{n \choose r-k}={m+n \choose r}$.
Does anyone know what happens if we walk bigger steps with k? Say we skip all the odd ks, is something like
$\sum_{k=0}^{r/2}{m \choose 2k}{n \choose r-2k}=\frac{1}{2} {m+n \choose r}$
or at least 
$\sum_{k=0}^{r/2}{m \choose 2k}{n \choose r-2k}=\Theta \left( \frac{1}{2} {m+n \choose r}\right)$
true? 
Maybe someone here has even some general insight on other step widths?
Thank you!

REPLY [3 votes]: We derive a binomial identity which shows the deviation of OPs sum from $\frac{1}{2}\binom{m+n}{r}$. It  is convenient to use the  coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\tag{1}
\end{align*}

We assume wlog $n\geq m$ and obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^{r/2}}&\color{blue}{\binom{m}{2k}\binom{n}{r-2k}}\\
&=\sum_{k\geq 0}\binom{m}{2k}[z^{r-2k}](1+z)^n\tag{2}\\
&=[z^r](1+z)^n\sum_{k\geq 0}\binom{m}{2k}z^{2k}\tag{3}\\
&=[z^r](1+z)^n\frac{1}{2}\left((1+z)^m+(1-z)^m\right)\tag{4}\\
&=\frac{1}{2}[z^r](1+z)^{m+n}+\frac{1}{2}[z^r](1+z)^n(1-z)^m\\
&=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}[z^r](1-z^2)^m(1+z)^{n-m}\tag{5}\\
&=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}[z^r]\sum_{k=0}^{r/2}\binom{m}{k}(-1)^kz^{2k}(1+z)^{n-m}\\
&=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}\sum_{k=0}^{r/2}\binom{m}{k}(-1)^k[z^{r-2k}](1+z)^{n-m}\tag{6}\\
&\,\,\color{blue}{=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}\sum_{k=0}^{r/2}\binom{m}{k}\binom{n-m}{r-2k}(-1)^k}\tag{7}
\end{align*}

Comment:

In (2) we apply the coefficient of operator as indicated in (1) and we set the upper limit of the sum to $\infty$ without changing anything since we are adding zeros only.
In (3) we use the linearity of the coefficient of operator.
In (4) we write the sum as polynomial in closed form.
In (5) we select the coefficient of $z^r$ of the left polynomial and we rewrite the other polynomial keeping in mind that $n\geq m$.
In (6) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
In (7) we select the coefficient of $z^{r-2k}$.