Let $n$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of $\frac{m}{n}$ is

$n:\ 5 \text{ boys},\ 5 \text{ girls},\ \text{all } 5 \text{ girls consecutive}$ $m:\ 5 \text{ boys},\ 5 \text{ girls},\ \text{exactly } 4 \text{ girls consecutive}$ $\text{To find: } \dfrac{m}{n}$ $\text{Plan: } 1.\ n;\quad 2.\ m\ (\text{runs } 4+1);\quad 3.\ m/n$ $\text{Glue all 5 girls into one block: } 5 + 1 = 6 \text{ units}$ $n = 6! \times 5!$ $n = 6! \times 5! = 720 \times 120 = 86400$ $\text{Choose 4 of 5 girls and arrange: } \binom{5}{4} \times 4! = 5 \times 24 = 120$ $\text{Exactly 4} \Rightarrow \text{girls form two runs: } 4 + 1$ $\text{the single girl must not touch the 4-block}$ $5! = 120 \Rightarrow 6 \text{ gaps}$ $6 \times 5 = 30 \text{ ways to place the block and the single girl}$ $m = 5! \times \binom{5}{4} \times 4! \times (6 \times 5)$ $m = 120 \times 120 \times 30 = 432000$ $\dfrac{m}{n} = \dfrac{432000}{86400} = 5$ $\boxed{5}$