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

Hi Menka,

We have to select according to the given condition....(selection three people would be same as placing three people with n-3 people already sitting in a circle)

Consider..... (n-3) peopple in a straight line, like this (where the crosses depict people)

    x x x x x x x ...... x   ------------(n-3 crosses)

Now there are n-2 gaps here, where 3 people can be placed in ^n-2C₃ ways (so that they are not adjacent)

Now if we place people in the 1st place before the 1st x and in the last place after the last x, in a circle they will be adjacent (so this case has to be removed)

The no of arrangements in which 1 person is in the 1st place, and another person in the last place is n-4.

So the favourable cases is ^n-2C₃ - (n-4) = n(n-4)(n-5)/6.

Hope it helps.

Bestg Regards,

Ashwin (IIT MadraS).

Approved