Question

In how many ways $10$ boys and $5$ girls sit around a circular table so that no two girls sit together?

• A. $\frac{9!\times 10!}{5!}$
• B. $9!\times 10!$
• C. $14!$
• D. $15!$

Answer 1

The correct option is A $\frac{9!\times 10!}{5!}$

First let us seat the $10$ boys, this can be done in $(10-1)!=9!$

Now there are $10$ gaps ( $1$ between each pair of boys ) in which $5$ girls are to be seated , this can be done in ${}^{10}{P}_5$ ways.

$\Rightarrow$ Total $=9!\times {}^{10}{P}_5=\frac{9!\times 10!}{5!}$

Hence, the answer is $\frac{9!\times 10!}{5!}.$