The number of ways of choosing triplet (x, y, z) such that z x, y and x, y, z 1,2, , n, n+1 is

Step-by-step solution

When z=n+1 we can choose x, y from 1,2, , n when z = n +1, x , y can be chosen in n ^2 ways and z = n , x , y can be chosen in ( n -1)^2 ways and so on n^2+(n-1)^2+ +1^2= 1 6 n(n+1)(2 n+1) ways of choosing triplets Alternatively triplets with x = y z , x y z , y x z can be chosen in ^ n+1 C₂, ^ n+1 C₃, ^ n+1 C₃ ways. There are ^ n+1 C₂+2 ( ^ n+1 C₃ )= ^ n+2 C₂+ ^ n+1 C₃=2 ( ^ n+2 C₃ )- ^ n+1 C₂