Question

$6$ boys and $4$ girls are to be arranged in a row. Let $m$ be the number of arrangements in which no two girls are together. Let $n$ denote the number of arrangements in which exactly two boys are together. Then the value of $\frac{m}{n}$ is

Answer 1

$\_\_B_1\_\_B_2\_\_B_3\_\_B_4\_\_B_5\_\_B_6\_\_$
$6$ boys can be arranged in $6!$ ways.
$4$ girls can be arranged in these $7$ blank spaces in ${}^7{C}_4\times 4!$ ways.
$\therefore m=6!\times {}^7{C}_4\times 4!$

$\_\_G_1\_\_G_2\_\_G_3\_\_G_4\_\_$
$4$ girls can be arranged in $4!$ ways.
$2$ boys can be selected in ${}^6{C}_2$ ways.
$6$ boys can be arranged in these $5$ blank spaces in $5!\times {}^6{C}_2\times 2$ ways.
$\therefore n=5!\times {}^6{C}_2\times 2\times 4!$

Hence, $\frac{m}{n}=\frac{6!\times {}^7{C}_4\times 4!}{5!\times {}^6{C}_2\times 2\times 4!}=7$