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in Complex number and Quadratic equations by (46.6k points)

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

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Let the n things be x1,x2,....xn.

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, ....xn-2xn - 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

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