All possible groups of 3 distinct numbers from among A, B, C, D and E are formed. If the aggregate of sums of numbers of each group is 120, then what is the arithmetic mean of A, B, C, D and E ?

Explanation

There are \binom{5}{3} = 10 groups of 3. Each number appears in \binom{4}{2} = 6 groups. The total aggregate sum is 6(A+B+C+D+E) = 120, giving A+B+C+D+E = 20. The mean of the 5 numbers is 20 / 5 = 4.

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