Question

If n distinct objects are arranged in a circle, show that the number of ways of selecting three of these n things so that no two of them is next to each other is n(n - 4)(n - 5)/6

Answer 1

Let the n things be x₁,x₂,....x_n.

The first choice may be one of these n things, and this is done in ⁿC₁ ways.

Suppose x₁ is the one chosen. The next two may be chosen – excluding x₁ and, the two next to x₁ , namely, x₂ , xn from the remaining (n – 3) in n−3C₂ ways. Of these ⁿ⁻³C₂ there are (n – 4) selections when the second two chosen are next to each other, like x₃x₄,x₄x₅, ....x_n-2x_{n - 1.} 

Therefore, the number of ways of selecting the second two after x₁ is chosen, so that the two are not next to each other is

The two objects can be relatively interchanged in two ways. Further, the order of the choice of the three is not to be considered. Hence, the number of ways of choice of the three is