Question

Let "k" be the number of ways $12$ balls be divided between $2$ boys, one receiving $5$ and the other $7$ balls and let "m" be the number of ways these $12$ balls be divided into groups of $5$, $4$ and $3$ balls respectively.
find $2$( $\frac{m}{k}$) ?

Answer 1

$\left(i\right)$
First boy can be given $5$ balls out $12$ balls in ${}^{12}{C}_5$ ways. Second boy can be given $7$ balls out of the remaining $7$ balls in ${}^7{C}_7$. But there order is important, boys interchange by $2$ types. Then the required number of ways
$={}^{12}{C}_5{\times }^7{C}_7\times 2!$
$=\frac{12!}{5!7!}\times 1\times 2!$
$=\frac{12!2}{5!7!}$
$=\frac{\mathrm{12.11.10.9.8.7}!.2}{\mathrm{5.4.3.2.1.}7!}=1584$
$\left(ii\right)$
First group can be given $5$ balls out of $12$ balls in ${}^{12}{C}_5$ ways. Second group can be given $4$ balls out of the remaining $7$ balls in ${}^7{C}_4$ ways and $3$ balls can be given out of the remaining $3$ balls in ${}^3{C}_3$ ways.
Hence the required number of ways (Here order of groups are not important)
$={}^{12}{C}_5{\times }^7{C}_4{\times }^3{C}_3$
$=\frac{12!}{5!7!}\times \frac{7!}{4!3!}\times 1=\frac{12!}{5!4!3!}=27720$