If there are 6 girls and 5 boys who sit in a row, then the probability that no two boys sit together is:

$\frac{6!6!}{2!11!}$ $\frac{7!5!}{2!11!}$ $\frac{6!7!}{2!11!}$ None of these

$\frac{6!7!}{2!11!}$

⇒ Required probability = $\frac{{^7C}_5×5!×6!}{11!}=\frac{7!\,6!}{11!\,2!}$