How to find $\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}$? Problem was to find $$\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}.$$ My partial progress i tried to motivate such that upper term in binomial terms gets constant rather than variable , so i thought of changing it to form $\binom{n}{n-r}\binom{n-r}{n-2r}$. Now it's of the form $\binom{n}{m}\binom{m}{k}$. We know it is equivalent to $\binom{n}{k}\binom{n-k}{m-k}$. Doing this we get $\binom{n}{2r}\binom{2r}{r}2^{n-2r}$. Now what I need to do as next step as such still nothing the upper or lower terms r any constant values(sum to ) , all are variables still. Now by one answer by Robpratt i got by snake oil , is there any other way to solve this problem ?

Applying
$$\binom{n}{k}\binom{n-k}{m-k}=\binom{n}{m}\binom{m}{k}$$
with $m=2r$ and $k=r$, we have
$$\binom{n}{r}\binom{n-r}{r} = \binom{n}{2r}\binom{2r}{r}.$$
Now snake oil yields
\begin{align}
\sum_{n\ge 0}\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r} z^n
&=\sum_{n\ge 0}\sum_{r\ge 0} \binom{n}{2r}\binom{2r}{r} 2^{n-2r} z^n \\
&=\sum_{r\ge 0}\binom{2r}{r} 2^{-2r} \sum_{n\ge 2r} \binom{n}{2r} (2z)^n \\
&=\sum_{r\ge 0}\binom{2r}{r} 2^{-2r} \frac{(2z)^{2r}}{(1-2z)^{2r+1}} \\
&=\frac{1}{1-2z}\sum_{r\ge 0}\binom{2r}{r} \left[\left(\frac{z}{1-2z}\right)^2\right]^r \\
&=\frac{1}{1-2z}\cdot\frac{1}{\sqrt{1-4\left(\frac{z}{1-2z}\right)^2}} \\
&=\frac{1}{\sqrt{(1-2z)^2-4z^2}} \\
&=\frac{1}{\sqrt{1-4z}} \\
&=\sum_{n\ge 0} \binom{2n}{n}z^n.
\end{align}
So $$\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r} = \binom{2n}{n}.$$
