Find the arithmetic mean of first 6 natural numbers.

Explanation

Obviously the smallest elements in these triplets are 1,2,3 and 4. We shall find the number of triplets `n(1),n(2),n(3),n(4)` respectively having 1,2,3, and 4 as the smallest elements. For 1 to be smallest element, we have to select any 2 elements from the remaining 5 elements 2,3,4,5,6, so that `n(1)=""^(5)C_(2)` . For 2 to be the smallest element, we have to select 2 elements from 4 higher numbers 3,4,5,6 in `""^(4)C_(2)` ways, so that `n(2)=""^(4)C_(2)` . Similarly `n(3)=""^(3)C_(2)` and `n(4)=""^(2)C_(2)`.
Augmented sum of all smallest numbers.
`=1xx""^(5)C_(2)+2 xx ""_(4)C_(2)+3xx ""_(3)C_(2)+4xx""_(2)C_(2)`
`=10+12+9+4`
=35
and the total number of these smallest number
`=""^(5)C_(2)+""^(4)C_(2)+""^(3)C_(2)+""^(2)C_(2)`
`=10+6+3+1=20`
Hence the A.M. required`=(35)/(20)=(7)/(4)`