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. $\frac{\left(4!\right)\left(4!\right)}{2!}$

Answer 1

Answer: (B)

First we fix the alternate position of girls and they can be arranged in $4!$ ways and in the five places five boys can be arranged in $^{5}P_{5}$ ways.
$\therefore $ Total number of ways $= 4! \times ^{5}P_{5} = 4! \times5!$