Combinatorics

Let X be the sum of the numbers selected. Then, E(X)=\sum_{i}^{ }x_{i}P(X=x_{i}). Now, P(X=x_{i})=\frac{1}{\binom{10}{3}}(as the numbers are selected uniformly without replacement). \sum_{i}^{ }x_{i}is the sum of the sums of all possible triplets. The sum contains \binom{9}{2} \frac{10}{9}'s, \binom{9}{2} \frac{20}{9}'s and so on. \therefore E(X)=\frac{1}{\binom{10}{3}}[\binom{9}{2}\frac{10}{9}+\binom{9}{2}\frac{20}{9}+...+\binom{9}{2}\frac{5120}{9}] =\frac{1}{\binom{10}{3}}\binom{9}{2}\frac{10}{9}[\frac{2^{10}-1}{2-1}] =\frac{\binom{9}{2}}{\binom{10}{3}}\frac{10}{9}\times 1023.