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 ?
 A: 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 $$\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r} = \binom{2n}{n}.$$
A: In seeking to evaluate
$$\sum_{r\ge 0} {n\choose r} {n-r\choose r} 2^{n-2r}$$
we write
$$\sum_{r\ge 0} {n\choose r} {n-r\choose n-2r} 2^{n-2r}
= 2^n [z^n] (1+z)^n
\sum_{r\ge 0} {n\choose r} 2^{-2r} \frac{z^{2r}}{(1+z)^r}
\\ = 2^n [z^n] (1+z)^n
\left(1+\frac{z^2}{4(1+z)}\right)^n
= \frac{1}{2^n} [z^n] (2+z)^{2n}
= {2n\choose n}$$
which is the desired closed form. Note that ${n-r\choose r}$ is not zero when $r\gt n$ while the coefficient extractor on ${n-r\choose n-2r}$ is. We nonetheless have equality because we get zero from ${n\choose r}$ on the extra range $r\gt n.$ Both forms are zero when $\lfloor n/2\rfloor \lt r \le n.$
A: For a combinatorial proof of
$$\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r} = \binom{2n}{n},$$
first rewrite the LHS as
$$\sum_{r\ge 0} \binom{n}{2r}\binom{2r}{r} 2^{n-2r}$$ as in my other answer.  Both sides of the identity count the number of $n$-subsets of $\{1,\dots,2n\}$.  The RHS is clear.  For the LHS, condition on the number $r$ of "complementary" pairs $\{i,2n+1-i\}$ that appear in the $n$-subset.  Then there are $n-2r$ "singletons" where one element appears and $r$ complementary pairs where neither element appears.  This idea is explained in Proofs That Really Count: The Art of Combinatorial Proof (Identity 161).
A: $$
\begin{align}
\sum_{r=0}^n\binom{n}{r}\binom{n-r}{r}2^{n-2r}
&=\sum_{r=0}^n\binom{2r}{r}\binom{n}{2r}2^{n-2r}\tag1\\
&=\frac{2^n}{2\pi}\sum_{r=0}^n\int_0^{2\pi}\cos^{2r}(x)\binom{n}{2r}\,\mathrm{d}x\tag2\\
&=\frac{2^n}{4\pi}\int_0^{2\pi}\left((1+\cos(x))^n+(1-\cos(x))^n\right)\mathrm{d}x\tag3\\
&=\frac{2^n}{4\pi}\int_0^{2\pi}\left(2^n\cos^{2n}(x/2)+2^n\sin^{2n}(x/2)\right)\mathrm{d}x\tag4\\
&=\frac{4^n}{2\pi}\int_0^{2\pi}\cos^{2n}(x/2)\,\mathrm{d}x\tag5\\
&=\binom{2n}{n}\tag6
\end{align}
$$
Explanation:
$(1)$: $\binom{n}{r}\binom{n-r}{r}=\binom{2r}{r}\binom{n}{2r}$
$(2)$: $\frac1{2\pi}\int_0^{2\pi}\cos^{2r}(x)\,\mathrm{d}x=4^{-r}\binom{2r}{r}$
$(3)$: $\sum_{r=0}^n\binom{n}{2r}x^{2r}=\frac12\left((1+x)^n+(1-x)^n\right)$
$(4)$: $1+\cos(x)=2\cos^2(x/2)$ and $1-\cos(x)=2\sin^2(x/2)$
$(5)$: substitute $x\mapsto x+\pi$ in the $\sin^{2n}(x/2)$ integral
$(6)$: substitute $x\mapsto2x$ then
$\phantom{\text{(6):}}$ $\frac1{\pi}\int_0^{\pi}\cos^{2n}(x)\,\mathrm{d}x=4^{-n}\binom{2n}{n}$
