Question

A gentleman invites $13$ guests to a dinner and places $8$ of them at one table and remaining $5$ at the other, the tables being round. The number of ways he can arrange the guests is

• A. $\frac{11!}{40}$
• B. $9!$
• C. $\frac{12!}{40}$
• D. $\frac{13!}{40}$

Answer 1

Answer: (D)

$\frac{13!}{40}$

The number of ways in which $13$ guest may be divided into groups of $8$ and $5{=}^{13}{C}_5=\frac{13!}{5!8!}$.

Now, corresponding to one such group, the $8$ guests may be seated at one round table in $(8-1)!\Rightarrow 7!$ ways

and the five guests at the other table in $(5-1)!=4!$ ways.

But each way of arranging the first group of $8$ persons can be associated with each way of arranging the second group of $5$,

Therefore, the two processes can be performed together in $7!\times 4!$ ways.

Hence required number of arrangements

$=\frac{13!}{5!8!}\times 7!\times 4!=\frac{13!}{5.4!\times 8.7!}\times 7!\times 4!=\frac{13!}{40}$