# Question 5 & 6 Review Exercise 6

Solutions of Question 5 & 6 of Review Exercise 6 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 5(i)

In how many different ways can be six children seated at a round table if certain two students refuse to sit next to each other?

### Solution

The total number of seats are six so $$n=6$$

The total different arrangements round a circular table are $$(n-1) !=(6-1) !=5 !=120$$

If two students refuse to sit next to each other, then the total possible arrangements are: $120-24=96$

## Question 5(ii)

In how many different ways can be six children seated at a round table if certain two students insist on sitting next to each other?

### Solution

The total number of seats are six so $n=6$.

The total different arrangements round a circular table are $(n-1) !=(6-1) !=5 !=120$.

If two students insist to seat next to each, we shall treat the two seats like one seat, and hence the total number of arrangements round the circle in this case are $$(n-1) !=(5-1) !=4 !=24$$

## Question 6

Six people including Faisal and Saima are to be seated at a round circular table. Find the probability that Faisal and Saima are seated next to each other

### Solution

The total number of ways sitting of six people around a circular table are: $$(n-1) !=(6-1) !=5 !=120$$ Is Faisal Saima will sit next to each other, then we shall treat both of them as single.

In this case the total number of ways to give seat to these six people are: $$4 ! \cdot 2 !=48$$

Because the five people will permute by $(5-1) !$ and Faisal and Saima can permute by 2 !

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