Question

The number of ways of choosing triplet (x, y, z) such that z ≥ max{x,y} and x,y,z ∈ {1,2, ....,n,.n + 1} is

(A)   ^{n + 1}C₃ +  ^{n + 2}C₃

(B)  1/6n(n +)(2n + 1)

(C) 1² + 2² + ....+ n²

(D)  2(^{n + 2}C₃) - ^{n + 1}C₂

Answer 1

Correct option  (B)  1/6n(n +)(2n + 1)

When z = n + 1 we can choose x, y from {1, 2, …, n}

Therefore, when z = n + 1; x, y can be chosen in n² ways and z = n; x, y can be chosen in (n – 1)² ways and so on Therefore,

n² + (n + 1)² + ...+ 1² = 1/6n(n + 1)(2n + 1) ways of choosing triplets.