Question

A number of 18 guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side. Determine the number of ways in which the sitting arrangements can be made.

Answer 1

Hint: We will first start by making the person who has to sit on a particular side sit and to find the remaining seats on each side and the person in total and finally we will permute the remaining persons on each side.

Complete step-by-step answer:
Now, we have been given that 18 guests have to be seated half on each side of a long table. Now, four particular guests want to sit on one particular side and three others on the other side.
Now, we have the person on each side $=\dfrac{18}{2}=9$.
Now, the persons who has their side and seat fixed as,
$\begin{aligned} & 18-4-3 \\\ & \Rightarrow 18-7=11 \\\ \end{aligned}$
Now, since 4 people are already seated on one side and 3 on the other side. So, we have 5 and 6 seats remaining on each side.
Now, we know the ways of selecting r objects out of n is ${}^{n}{{C}_{r}}$. Also, ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$.
Now, we can choose 6 men’s for one side among 11 men’s in ${}^{11}{{C}_{6}}$ ways. Also, we can choose the remaining 5 men for the other side in ${}^{5}{{C}_{5}}$ ways.
Now, we can permute the 9 men among themselves in 9! Ways on each side.
So, by multiplying all the possible ways, we have the number of ways in which sitting arrangement can be made is,
${}^{11}{{C}_{6}}\times {}^{5}{{C}_{5}}\times 9!\times 9!$

Note: To solve this question we have used the fact that the ways of selecting r objects among n is ${}^{n}{{C}_{r}}$. Also, it is important to remember that we have multiplied the ways by 9! two times because we can permute 9 men on each side in 9! ways each.