<s>[INST] Evaluation of $\sum_{r=1}^{n}r\cdot (r-1)\cdot \binom{n}{r} = $ Evaluation of $\mathop{\displaystyle \sum_{r=1}^{n}r\cdot (r-1)\cdot \binom{n}{r} = }$
$\bf{My\; Try::}$ Given $$\displaystyle \sum_{r=1}^{n}r\cdot (r-1)\cdot \binom{n}{r}\;,$$ Now Using the formula $$\displaystyle \binom{n}{r} = \frac{n}{r}\cdot \binom{n-1}{r-1}.$$ 
$$\mathop{\displaystyle =\sum_{r=1}^{n}r\cdot (r-1)\cdot \frac{n}{r}\cdot \binom{n-1}{r-1}}$$
$$\mathop{\displaystyle  = n\sum_{r=2}^{n}(r-1)\cdot \frac{n-1}{r-1}\cdot \binom{n-2}{r-2} = n\cdot (n-1)\sum_{r=2}^{n}\binom{n-2}{r-2}}$$
$$\mathop{\displaystyle  = n\cdot (n-1)\cdot \left[\binom{n-2}{0}+\binom{n-2}{1}+..........+\binom{n-2}{n-2}\right] = n(n-1)\cdot 2^{n-2}}$$
My question is can we solve it using combinatorial argument, If yes then plz explain here
Thanks
 [/INST] $\sum_{r=0}^{n} r(r-1)\binom{n}{r} = 2 \sum_{r=0}^n \binom{n}{r} \binom{r}{2}$. The summand is the number of ways of choosing from $n$ objects an $r$-tuple and a 2-subtuple, so the right-hand sum is just the number of ways of choosing a 2-tuple multiplied by the number of possible tuples the 2-subtuple could have been drawn from. That is, the number of 2-tuples multiplied by $2^{n-2}$, or $\binom{n}{2} \times 2^{n-2}$.
Therefore, the left-hand sum is $n(n-1) \times 2^{n-2}$.
To expand on the $2^{n-2}$: the set of tuples from which a given 2-subtuple $(a,b)$ has been drawn, bijects with the set of tuples drawn from a set of size $n-2$, by the bijection "remove $a$ and $b$ from the tuple".
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