product of binomial coefficients taken some or all at atime If $\binom{n}{0} , \binom{n}{1} , \binom{n}{2} , \binom{n}{2} .... \binom{n}{n}$ denote the binomial coefficients in the expansion of ${(1+1)}^n$ then what is the difference between 
$$\sum_{r=0}^{n} \sum_{s=0}^{n} \binom{n}{r}\binom{n}{s}$$
and 
$$\sum \sum \binom{n}{r}\binom{n}{s}$$ 
$r$ is greater than or equal to $0$ but it is less than $s$ and $s$ is less than or equal to $n$
How can we understand the differences in the meaning of the questions ?
 A: The products that are in the double summations can be featured in the following double array (for $n=4$):
$$\begin{array}{|l||l|l|l|l|}
\hline
&\color{red}{\binom{4}{0}}&\color{red}{\binom{4}{1}}&\color{red}{\binom{4}{2}}&\color{red}{\binom{4}{3}}&\color{red}{\binom{4}{4}}\\ 
\hline
\color{red}{\binom{4}{0}}&&*&*&*&*\\
\hline
\color{red}{\binom{4}{1}}&&&*&*&*\\
\hline
\color{red}{\binom{4}{2}}&&&&*&*\\
\hline
\color{red}{\binom{4}{3}}&&&&&*\\
\hline
\color{red}{\binom{4}{4}}&&&&&\\
\hline
\end{array}$$
The first sum uses all the boxes. The second sum, assumed to be 
$$\sum \sum \binom{n}{r}\binom{n}{s} \  \text{for} \ 0 \leq r < s\leq n$$
uses only the starred boxes.
A: The following representations of the double sums might be useful.

We have
  \begin{align*}
\sum_{r=0}^n\sum_{s=0}^n\binom{n}{r}\binom{n}{s}&=\sum_{\color{blue}{0\leq r,s\leq n}}\binom{n}{r}\binom{n}{s}
=\sum_{s=0}^n\sum_{r=0}^n\binom{n}{r}\binom{n}{s}\\
\sum_{r=0}^{n-1}\sum_{s=r+1}^n\binom{n}{r}\binom{n}{s}&=\sum_{\color{blue}{0\leq r<s\leq n}}\binom{n}{r}\binom{n}{s}
=\sum_{s=1}^n\sum_{r=0}^{s-1}\binom{n}{r}\binom{n}{s}\
\end{align*}

A: $$\sum_{r=0}^n\sum_{s=0}^n{n\choose r}{n\choose s}=2\sum_{r=0}^{n-1}\sum_{s=r+1}^n{n\choose r}{n\choose s}+\sum_{t=0}^n{n\choose t}^2$$
