Binomial Summations 1

I think this question is pretty easy and over-rated. Let's see how to do it!

Take $\displaystyle \mathop{\sum\sum}_{0\le i<j\le n} i\binom{n}{j}$

Now, for a particular value of $j$ , $\begin{aligned}\sum_{i=0}^{j-1}i\binom nj&=\binom nj\sum_0^{j-1}i\\&=\dfrac{j(j-1)}2\binom nj \end{aligned}$

So, $\begin{aligned}\displaystyle \mathop{\sum\sum}_{0\le i<j\le n} i\binom{n}{j} &= \sum_{j=1}^n \dfrac{j(j-1)}2\binom nj\\&=\dfrac{n(n-1)}2\sum_{j=1}^n \binom{n-2}{j-2}\\&=\dfrac{n(n-1)}2\sum_{j=0}^{n-2} \binom{n-2}{j}\\&=\Large\displaystyle\boxed{\dfrac{n(n-1)}2 2^{n-2}} \end{aligned}$

$\displaystyle\Huge\therefore \boxed{\dfrac{12(12-1)}2 2^{10} = 67584}$