Double Sigma combinations: $\sum_{j=0}^{11}\sum_{i=j}^{11}\binom ij$ Find the sum.
$$\sum_{j=0}^{11}\sum_{i=j}^{11}{i \choose j}.$$
I am getting the answer as $4095$ although the answer is given as $4092$.
 A: Say you want to choose a non-empty subset of $\{0,1,2,\ldots,11\}$, which has $12$ members.
Since the subset to be chosen is non-empty, it has a largest member.  Let $j$ be the number of members other than the largest member. There can be $11$ other members or $0$ other members or anything in between. Call the largest member $i$, and observe that $i$ must be at least $j$, since it can be equal to $j$ if, but only if, the $j$ non-largest members are $0,1,2,\ldots,j-1.$ The largest member is in the set $\{j,j+1,j+2,\ldots,11\}.$ Thus the total number of possibilities is
$$
\sum_{j=0}^{11} \sum_{i=j}^{11} \Big( \text{number of ways to choose $j$ members from $\{0,1,2,\ldots,i-1\}$ } \Big).
$$
$$
= \sum_{j=0}^{11} \sum_{i=j}^{11} \binom i j.
$$
But the total number of non-empty subsets of $\{0,1,2,\ldots,11\}$ is $2^{12}-1 = 4095.$
A: 
Your answer is correct, since
  \begin{align*}
\sum_{j=0}^{11}\sum_{i=j}^{11}\binom{i}{j}&=\sum_{0\leq j\leq i\leq 11}\binom{i}{j}\\
&=\sum_{i=0}^{11}\sum_{j=0}^i\binom{i}{j}\\
&=\sum_{i=0}^{11}2^i\\
&=\frac{2^{12}-1}{2-1}\\
&=4095
\end{align*}

