====== 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.
=====Question 10(i)=====
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?
====Solution====
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}
=====Question 10(ii)=====
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?
====Solution====
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}
====Go To====
[[math-11-kpk:sol:unit06:ex6-2-p5 |< Question 9 ]]
[[math-11-kpk:sol:unit06:ex6-2-p7|Question 11 >]]