Question

6 girls and 5 boys sit together randomly in a row, the probability that no two boys sit together, is

• A. $\frac{6!5!}{11!}$
• B. $\frac{6!6!}{11!}$
• C. $\frac{6!7!}{2!11!}$
• D. $\frac{5!7!}{2!11!}$

Answer 1

The correct option is C $\frac{6!7!}{2!11!}$

First arrange 6 girls in 6! ways
then Places for boys = 7 (Here, 6 girls and 5 boys)
Hence, probability no two boys sit together
$=\frac{{}^7{P}_5\times 6!}{11!}=\frac{7!6!}{2!11!}$