Let $S=\{1, 2, 3,\cdots , n\}$. If X denote the set of all subsets of S containing exactly two elements, then the value of $\sum _{A\in X}$ (min A) is

There are exactly ($n – 1$) subsets of S containing two elements having 1 as least element; exactly $(n – 2)$

subsets of S having 2 as least element and so on. Thus,

$\sum _{A=X} min\left(A\right)=1(n-1)+2(n-2)+\cdots +(n-1) \left(1\right)$

$\begin{array}{l}=\sum _{r=1}^{n-1} r(n-r)=n\sum _{r=1}^{n-1} r-\sum _{r=1}^{n-1} r^2\\ =n\left\{\frac{1}{2}n(n-1)\right\}-\frac{1}{6}n(n-1)(2n-1)\\ =\frac{1}{6}n(n-1)\{3n-2n+1\}\\ =\frac{1}{6}(n+1)n(n-1){=}^{n+1}{C}_3\end{array}$