Question

In how many ways can $5$ boys and $5$ girls sit in a circle so that no two boys sit together?

• A. $5! \times 5!$
• B. $4! \times 5!$
• C. $\frac{5! \times 5!}{2}$
• D. None of these

Answer 1

Answer: (B)

$4! \times 5!$

Answer 2

First we fix the alternate position of the girls.
Five girls can be seated around the circle in $(5-1) !=4 !, 5$ boys can be seated in five -vacant place by $5 !$
$\therefore$ Required number of ways $=4 ! \times 5 !$