Question

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

Answer 1

$\text{Case}:1n=5!\times 6!$

$\text{where}5!=\text{arranging of five girls}$

$\text{and}6!=\text{arranging of five boys and five consecutive girls}$
$\text{Case}:2$
$\_\_B\_\_B\_\_B\_\_B\_\_B\_\_$
$m=5!\times {}^5{C}_4\times 4!\times {}^6{C}_2\times 2!$
$\text{where}5!=\text{arranging of five boys,}$
${}^5{C}_4=\text{selecting of four girls,}$\
$4!=\text{arranging of four girls,}$
${}^6{C}_2=\text{selecting two gaps in between five boys,}$
$\text{and}2!=\text{arranging of four consecutive girls and one single girl}$
$\therefore \frac{m}{n}=\frac{5!\times {}^5{C}_4\times 4!\times {}^6{C}_2\times 2!}{5!\times 6!}=5$