Ans: $\dfrac{5}{42}$
$P(\text{no boys are next to each other in the queue})$
$P(\text{no boys are next to each other in the queue})$
$\begin{array}{cl}
= & \dfrac{5! \times P^6_4}{9!} \\
= & \dfrac{5}{42}
\end{array}$
