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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).

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).

Grade:12th Pass

1 Answers

Ashwin Muralidharan IIT Madras
290 Points
10 years ago

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).

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