proof of binomial identity involving double sum It's asked to simplify the sum
$$\displaystyle\sum_{0\le i <j\le n+1} \binom{n}{i} \times \binom{n+1}{j} $$
inspecting first values of $n$ shows the sum if apparently equal to $4^n$
I tried re-writing the sum as 
$$\displaystyle\sum_{j=1}^{n+1} \binom{n+1}{j} \displaystyle\sum_{i=0}^{j-1} \binom{n}{i}  $$ but that doesn't seem to lead us to the result.
Any suggestions are welcome.
Thanks.

Hint: For every $(i,j)$ with $0\leq i\leq n$, $0\leq j\leq n+1$ we have either $i<j$ or $n-i<n+1-j$ but not both.  So
$$
\sum_{0\leq i<j\leq n+1}\binom{n}{i}\binom{n+1}{j}
=\sum_{i,j}\binom{n}{i}\binom{n+1}{j}-\sum_{\substack{n-i<n+1-j\\0\leq i\leq n\\0\leq j\leq n-1}}\binom{n}{i}\binom{n+1}{j}
$$
and now use $\binom{n}{i}=\binom{n}{n-i}$, $\binom{n+1}{j}=\binom{n+1}{n+1-j}$.
