<s>[INST] Half of Vandermonde's Identity 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!
 [/INST] 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 $z^{r-2k}$.
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