Question

In how many ways fifteen different items my be given to $A,B,C$ such that$A$ gets $3$, $B$ gets $5$ and remaining goes to $C$.

Answer 1

There are $15$ items, $A$ gets $3,B$ gets $5,C$ gets $(15-3-5)=7$

$A$ may get any $3$ of the $15$

$={}^{15}{C}_3$

$B$ may get any $5$ of the remaining $12$

$={}^{12}{C}_5$

$C$ gets remaining $7$

$={}^7{C}_7$

$\Rightarrow$ Total $={}^{15}{C}_3\times {}^{12}{C}_5\times {}^7{C}_7$

$=\frac{15!\times 5!}{3!7!}$

Hence, the answer is $\frac{15!\times 5!}{3!7!}.$