Ans: (a) $3\ 628\ 800$ (b) $\dfrac{7}{15}$
- The required number of queues
$\begin{array}{cl}
= & (7 +3)! \\
= & 3\ 628\ 800
\end{array}$ - The required probability
$\begin{array}{cl}
= & \dfrac{7! \times P^8_3}{3\ 628\ 800} \\
= & \dfrac{7}{15}
\end{array}$
