the sum of elements of all possible subset $B$ is 
Let $A = \{1,2,3,4,\cdots \cdots ,22\}$ and $B$ is a subset of $A$ having exactly $11$ elements.
Then the sum of elements of all possible subset $B$ is

$\bf{Attempt}$ Number of subset of $A$ having excatly $11$ elements is $\displaystyle \binom{22}{11}$
Now how can i find sum of elements of all subset of $B$, could some help me , thanks
 A: For each 11-element subset $B$ of $A$, its complement $A\setminus B$ also is an 11-element subset.  The elements of those two subsets sum up to $S = 22\times23/2 = 253$.  As there are $\binom{22}{11}$ 11-element subsets, we have an overall total of  $S\times\binom{22}{11} = 178474296$ (which counts each such subset exactly twice, once as a subset in its own right, and once as a complement) giving an answer of $178474296/2 = 89237148$.
[This is equivalent to the answer noted in a comment, which I saw after answering.]
A: The generating function for subsets of $N_n:=\{1,\dots,n\}$ weighted by sums is
$$ \sum_{S \subseteq N_n} x^{\#S}q^{\operatorname{sum}(S)} = (1 + qx)(1 + q^2x)\cdots(1 + q^{n}x) = \sum_{k = 0}^n q^{k(k+1)/2} \begin{bmatrix} n \\ k \end{bmatrix}_q x^k. $$
So if we set $n = 22$ and sum over subsets of $N_{22}$ with $11$ elements, we get
$$ \sum_{B} q^{\operatorname{sum}(B)} = q^{66}\begin{bmatrix} 22 \\ 11 \end{bmatrix}_q. $$
Now differentiate with respect to $q$ to get
$$ \sum_{B} \operatorname{sum}(B)q^{\operatorname{sum}(B) - 1} = \frac{d}{dq} \left( q^{66}\begin{bmatrix} 22 \\ 11 \end{bmatrix}_q \right). $$
Finally set $q = 1$ to obtain
$$ \sum_{B} \operatorname{sum}(B) = 89237148. $$
Where the last quantity was computed using Mathematica:
D[q^66 QBinomial[22,11,q]//FunctionExpand, q] /. q -> 1

You can also give this same command to WolframAlpha.
