- #1

- 3,812

- 92

## Homework Statement

There are n persons sitting round a table. Prove that the number of different ways in which 3 persons can be selected so that no two are neighbours is ##\frac{1}{6}n(n-4)(n-5)##.

## Homework Equations

## The Attempt at a Solution

I don't really know how to tackle this problem. I started with considering a simpler case first. Let the number of persons be 6.

We select one person from 6. Then according to given condition, we cannot select the two persons sitting just next to our selected person. So the other two must be selected from the remaining three. We have to subtract the cases when two nearby persons are selected when we selected two persons from the remaining three. Hence, number of cases=6C1*3C2-2=16, but this wrong because if we substitute 6 in the given formula (which we have to prove), it gives 2.

Any help is appreciated. Thanks! :)