Girls joining a ring of boys

Fourteen boys, Charlie among them, stand in a circle. Ten girls then join, each taking a spot in one of the gaps between neighbouring boys, with at most one girl per gap; all such placements are equally likely. What is the probability that Charlie still has a boy on both sides?

$\tfrac{6}{91}$. There are $14$ gaps and the girls occupy $10$ of them, so $\binom{14}{10}$ equally likely choices. Charlie keeps two boy neighbours only if both gaps beside him stay empty, so the girls must use 10 of the other 12 gaps: $\binom{12}{10}/\binom{14}{10}=\tfrac{66}{1001}=\tfrac6{91}$.