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

Since total number of ways in which boys can occupy any place is $(5 - 1)\mspace{6mu}! = 4\mspace{6mu}!$ and the 5 girls can be sit accordingly in $5\mspace{6mu}!$ ways.

Hence required number of ways are $4\mspace{6mu}!\mspace{6mu} \times 5\mspace{6mu}!$.