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.