Double Sigma combinations: $\sum_{j=0}^{11}\sum_{i=j}^{11}\binom ij$

Question

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 $4092$.

Answer 1

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*}

Answer 2

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 $2^{12}-1 = 4095.$