# Question 10 Exercise 6.2

Solutions of Question 10 of Exercise 6.2 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.

In how many ways can five students be seated in a row of eight seals if a certain two students insist of sitting next to each other?

Total number of seats are eight, so $n=8$.

Number of students are five so, $r=5$.

The total number of ways these five students can be seated are: \begin{align}^8 P_5&=\dfrac{8 !}{(8-5) !}\\ &=\dfrac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 !}{3 !}\\ &=6720\end{align} If certain two students insist to sit next to each other then these two students will be handled as a single students and the eights seats will be considered as 7.

In this case the total number of ways are: \begin{align}^2 P_2 \times^7 P_4&=2 \times \dfrac{7 !}{(7-4) !}\\ &=2 \times\dfrac{7.6 .5 .4 .3 !}{3 !}\\ &=1680\end{align}

In how many ways can five students be seated in a row of eight seals if a certain two students refuse to sit next to each other?

Total number of seats are eight, so $n=8$.

Number of students are five so, $r=5$.

The total number of ways these five students can be seated are: \begin{align}^8 P_5&=\dfrac{8 !}{(8-5) !}\\ &=\dfrac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 !}{3 !}\\ &=6720\end{align} If certain two students refuse to sit next to each other, then the total number ways sitting these students in a row are: \begin{align}^7 P_4&= \dfrac{7 !}{(7-4) !}\\ &=\dfrac{7.6 .5 .4 .3 !}{3 !}\\ &=840\\ \text{then}\quad &6720-840=5880\end{align}