Question

A tea party is arranged for $16$ people along two sides of a large table with $8$ chairs on each side. Four men want to sit on one particular side and two on the other side. The number of ways in which they can be seated is
A. $\dfrac{6!8!10!}{4!6!}$
B. $\dfrac{8!8!10!}{4!6!}$
C. $\dfrac{8!8!6!}{6!4!}$
D. None of these

Answer 1

Hint: We will first separate the two sides of the chairs and find the number of ways that $4$ persons can arrange on one side and the number of ways that $2$ persons sit on other side and number of ways that the remaining persons can sit. Multiply all the obtained values to get the result.

Complete step by step answer:
Given that, there are $16$ people for a tea party. There are two sides with $8$ chairs for the party. Here name the two sides as $A$ and $B$.
Now $4$ persons want to sit on a particular side. Let us assume that the $4$ persons sit on side $A$ then the number of ways of arranging $4$ persons in $8$ chairs is
$A={}^{8}{{\text{P}}_{4}}$
If $4$persons sit on side $A$ then, from given data the two persons will sit on side $B$. Now the number of ways of arranging $2$ persons in $8$ chairs is
$B={}^{8}{{\text{P}}_{2}}$
Now the remaining people are $16-\left( 4+2 \right)=10$ persons will sit on $16-\left( 4+2 \right)=10$ chairs in $10!$ ways.
Now required number of ways is
$\begin{align}
  & {}^{8}{{\text{P}}_{4}}\times {}^{8}{{\text{P}}_{2}}\times 10!=\dfrac{8!}{\left( 8-4 \right)!}\times \dfrac{8!}{\left( 8-2 \right)!}\times 10! \\
 & =\dfrac{8!8!10!}{4!6!}
\end{align}$

So, the correct answer is “Option B”.

Note: You should know some basic concept of permutations. We can arrange $n$ things in $m$ places in ${}^{m}{{\text{P}}_{n}}$ ways. The value of ${}^{m}{{\text{P}}_{n}}$ is$\dfrac{m!}{\left( m-n \right)!}$