Evaluate $ \mathop{\sum^{n}\sum^{n}}_{i=0\ i < j\ j=0} i \binom{n}{j}$

Question

Evaluate $$S=\mathop{\sum^{n}\sum^{n}}_{i=0,\: i < j,\: j=0} i \binom{n}{j}$$

My try:

I considered:

$$P=\mathop{\sum^{n}\sum^{n}}_{i=0,\: i < j,\: j=0} i \binom{n}{j}+j \binom{n}{i} \tag{1}$$

By Symmetry we have:

$$P=\mathop{\sum^{n}\sum^{n}}_{i=0,\: i > j,\: j=0} i \binom{n}{j}+j \binom{n}{i} \tag{2}$$

Letting $$Q=\mathop{\sum^{n}\sum^{n}}_{i=0,\: i = j,\: j=0} i \binom{n}{j}+j \binom{n}{i}$$

Hence:

$$Q=2\sum_{i=0}^{n}i \binom{n}{i}=2n 2^{n-1}=n2^n \tag{3}$$

Adding $(1)$,$(2)$ and $(3)$ we get:

$$2P+Q=\sum_{i=0}^{n}\sum_{j=0}^{n}\left(i\binom{n}{j}+j \binom{n}{i}\right)$$

We get:

$$2P+Q=2\left(\sum_{i=0}^{n}\sum_{j=0}^{n}\left(i\binom{n}{j}\right)\right)$$

$$2P+Q=2\left(\frac{n(n+1)}{2}\right)2^n=n^22^n+n2^n$$

$\implies$

$$P=n^22^{n-1}$$

With $P$ can we calculate $S$?

Answer 1

$S=\sum^n\sum^n_{i=0,j=0,i<j}i{n\choose j}=\sum_{j=1}^n{n\choose j}\sum_{i=0}^{j-1}i=\sum_{j=1}^n{n\choose j}\frac{j(j-1)}2=\sum_{j=2}^n\frac{n!}{j!(n-j)!}\frac{j(j-1)}2\\= \frac12\sum_{j=2}^n\frac{n!}{(j-2)!(n-j)!}=\frac{n(n-1)}2\sum_{j=2}^n\frac{(n-2)!}{(j-2)!(n-j)!}=\frac{n(n-1)}2\sum_{j=2}^n{n-2\choose j-2}=\frac{n(n-1)}2\sum_{k=0}^{n-2}{n-2\choose k}=\frac{n(n-1)}2 2^{n-2}=n(n-1)2^{n-3}.$