# 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.

**Question**

*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?

**Solution**

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 ^{n}C_{3}.

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 = ^{n}C_{3}

Total restricted ways = x

Therefore,

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

= ^{n}C_{3} - 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 = ^{n}C_{3} - x

= (n!/3!(n-3)!)- n(n-3)

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

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