Trick to solve seating arrangement problems

This tip to solve permutation and combinations problems was submitted by Ayush, a HashLearn tutor from IIT, Dhanbad.

In this tip, you will learn how to quickly solve one kind of seating arrangement problems.


n people are sitting on a round table. In how many ways can 3 (where 3 is lesser than n) of them be selected randomly so that no two are consecutive?


First of all, we’ll find the total number of ways m people can be selected at random from these n people, it will be equal to nC3.

Now we need to find ways of arranging these 3 people in such a way so that no two of them are consecutive. So, first we count the total ways of finding 3 people such that atleast any two are consecutive (the restricting condition) and assume it to be x.

So for now,
Total ways in which they can be seated = nC3
Total restricted ways = x

Our required answer will be = Total Ways - Total Unfavourable ways
= nC3 - x

Now our job is to calculate x.

We first fix P1 and P2 and the third person is varied from P3 to Pn-1. In this way we count all the ways in which 3 people are selected in which a minimum of 2 are consecutive. The total count comes out to be (n-3).

Now this time we fix P2 and P3 and the third person is varied from P4 to Pn. This time also total count comes out to be (n-3).
This process of pairing is continued till the pair Pn and P1, because after this the triplets starts repeating.

The total number of pairs formed in this process is n.
Hence the total number of ways = n(n-3)

This table has n rows and (n-3) columns. The total number of ways thus is equal to product of rows and columns which is n(n-3).
Hence x=n(n-3)

Our answer = nC3 - x
= (n!/3!(n-3)!)- n(n-3)

= (n(n-4)(n-5))/6

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