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 $\frac{\mathrm{m}}{\mathrm{n}}$ is

For n ${\mathrm{B}}_1,{\mathrm{B}}_2,{\mathrm{B}}_3,{\mathrm{B}}_4,{\mathrm{B}}_5,({\mathrm{G}}_1,{\mathrm{G}}_2,{\mathrm{G}}_3,{\mathrm{G}}_4,{\mathrm{G}}_5)$
Number of arrangements $\Rightarrow \mathrm{n}=5!\times 6!$
For m
First arrange 5 boys in 5! Ways
$\uparrow {\mathrm{B}}_1\uparrow {\mathrm{B}}_2\uparrow {\mathrm{B}}_3\uparrow {\mathrm{B}}_4\uparrow {\mathrm{B}}_5\uparrow$
Now, we have to arrange 5 girls in such a way that group of four girls and the fifth girl are arranged in any two of the six positions shown as arrows.
Two positions can be selected in ${ }^6{\mathrm{C}}_2$ ways.
Four girls can be selected in ${ }^5{\mathrm{C}}_4$ ways
Now, this group and the fifth girl can be arranged in selected two positions is 2! Ways
Also, four girls arrange among themselves in 4! Ways
Hence, number of arrangements $=\mathrm{m}=5!{\times }^6{\mathrm{C}}_2{\times }^5{\mathrm{C}}_4\times 2!\times 4!$
$$\begin{gathered} =5!\times 15\times 2\times 5! \\ \therefore \frac{\mathrm{m}}{\mathrm{n}}=\frac{5!\times 15\times 2\times 5!}{5!\times 6!}=5 \end{gathered}$$