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.

Let X be the sum of the selected 3 numbers.
Now, E(X)={summation}x_i.P(X=x_i)=110C3{sumn.}x_i.
In {sumn.}x_i, there are 9C2 109's, 9C2 209's and so on.
Therefore E(X)=110C39C2[109+209+...+51209]=9C210C3109[1+2+...+512]=9C210C3109x1023.