Double summation with restriction $\sum_{0\leq\ r \leq s\leq n}\sum \left ( {n\choose r}+{n\choose s} \right )^2$.
I can  not understand how to evaluate the condition of r<s in the summation ...$\sum_{r=0}^n {n\choose r}^2$ I understand would end up  being ${2n\choose n}$...so i thought of using that idea..but I need help in understanding what the notation in the summation means
 A: It means the sum over all pairs $(r,s)$ with $0\leq r\leq s\leq n$.  It's another way of writing $$\sum_{r=0}^n\sum_{s=r}^n\binom nr^2+\binom ns^2$$
As to evaluating the sum, count how many times $\binom nk^2$ occurs for each $0\leq k\leq n$.  It occurs once as the $r$ term for each $s>k$, once as the $s$ term for each $r<k$ and once when $r=s=k$.
Given $0\leq k \leq n$, the term $\binom nk^2$ occurs as the $s$ term $k$ times, once for each $0\leq r<k.$  It occurs as the $r$ term $n-k$ times, once for each $k<s\leq n$.  It also occurs twice when $r=k=s$, so it ccurs $(n+2)$ times in all.  That is, the given sum is $$(n+2)\sum_{k=0}^n\binom nk^2$$
You can also do this more formally by breaking up the sum into two sums.  You'll have to reverse the order of summation in one of them.  I think the argument above, as well as being simpler, casts more light on what is going on.
Just for completeness, the formal argument would be
$$\begin{align}\sum_{r=0}^n\sum_{s=r}^n\binom nr^2+\binom ns^2&=
\sum_{r=0}^n\sum_{s=r}^n\binom nr^2+\sum_{r=0}^n\sum_{s=r}^n\binom ns^2\\
&=\sum_{r=0}^n(n-r+1)\binom nr^2+\sum_{s=0}^n\sum_{r=0}^s\binom ns^2\\
&=\sum_{r=0}^n(n-r+1)\binom nr^2+\sum_{s=0}^n(s+1)\binom ns^2\\
&=\sum_{r=0}^n(n-r+1)\binom nr^2+\sum_{r=0}^n(r+1)\binom nr^2\\
&=\sum_{r=0}^n(n+2)\binom nr^2=(n+2)\sum_{r=0}^n\binom nr^2
\end{align}$$
A: Common notations to sum up over the index region $0\leq r\leq s\leq n$ are
\begin{align*}
\sum_{r=0}^n\sum_{s=r}^n a_{r,s}=\sum_{\color{blue}{0\leq r\leq s\leq n}}a_{r,s}=
\sum_{s=0}^n\sum_{r=0}^n a_{r,s}\tag{1}
\end{align*}
If you check in each sum lower and upper bounds of the indices $r$ and $s$ it might become plausible.

Using this notation we obtain
\begin{align*}
&\color{blue}{\sum_{0\leq\ r \leq s\leq n}}\color{blue}{\left(\binom{n}{r}+\binom{n}{s}\right)^2}\\
&\quad=\sum_{0\leq\ r \leq s\leq n}\left(\binom{n}{r}^2+2\binom{n}{r}\binom{n}{s}+\binom{n}{s}^2\right)\tag{2}\\
&\quad=\sum_{r=0}^n\sum_{s=r}^n\binom{n}{r}^2+2\sum_{0\leq r\leq s\leq n}\binom{n}{r}\binom{n}{s}\\
&\qquad\qquad+\sum_{s=0}^n\sum_{r=0}^s\binom{n}{s}^2\tag{3}\\
&\quad=\sum_{r=0}^n\binom{n}{r}^2\sum_{s=r}^n1+2\sum_{0\leq r\leq s\leq n}\binom{n}{r}\binom{n}{s}\\
&\qquad\qquad +\sum_{s=0}^n\binom{n}{s}^2\sum_{r=0}^s1\tag{4}\\
&\quad=(n+2)\sum_{r=0}^n\binom{n}{r}\binom{n}{n-r}+2\sum_{0\leq r\leq s\leq n}\binom{n}{r}\binom{n}{s}\tag{5}\\
&\,\,\color{blue}{\quad=(n+2)\binom{2n}{n}++2\sum_{0\leq r\leq s\leq n}\binom{n}{r}\binom{n}{s}}\tag{6}
\end{align*}
which is quite a nice identity.

Comment:

*

*In (2) we apply the binomial theorem.


*In (3) we separate the double sums. We also choose the index summation in the left-hand sum and right-hand sum so, that the binomials do not depend on the inner sum which makes simplifications somewhat easier.


*In (4) we factor out the binomial coefficients, so that the inner sums are constants.


*In (5) we collect the left-hand sum and right-hand sum of (4) (with implicitly replacing $r$ with $s$ in the right-hand sum). We also use the identity $\binom{p}{q}=\binom{p}{p-q}$ as preparation for applying the next step.


*In (6) we apply the Vandermonde identity.
Note: For small $n$ of $\sum_{0\leq r\leq s\leq n}\binom{n}{r}\binom{n}{s}$ we get the sequence
\begin{align*}
\{12,46,184,746,3\,040,12\,412,50\,704,\ldots\}
\end{align*}
which is not archived in OEIS. This is an indication that there is no closed formula available.
