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

$n=5!\times 6!$
For second arrangement, 5 boys can be made to stand in a row in 5! ways, creating 6 alternate space for 4 girls. A group of 4 girls can be selected in${}^5{C}_4$ways.
A group of 4 and single girl can be arranged at 2 places out of 6 in ${}^6{P}_2$ ways. Also 4 girls can arrange themselves in 4! ways.
$\therefore m=5!\times {}^6{P}_2\times {}^5{C}_4\times 4!\frac{m}{n}=\frac{5!\times 6\times 5\times 5\times 4!}{5!\times 6!}=5$