Question

No. of ways of distributing $(p+q+r)$ different things to $3$ persons so that one person gets $p$ things, $2^{nd}$ one $q$ things $3^{rd}$ person $r$ things is

• A. $\frac{(p+q+r)!}{p!q!r!}\times 3!$
• B. $\frac{(p+q+r)!}{p!q!r!}$
• C. $\frac{(p+q+r)!}{3!p!q!r!}$
• D. $\frac{(p+q+r)!}{3!pqr}$

Answer 1

The correct option is A $\frac{(p+q+r)!}{p!q!r!}\times 3!$

$1^{st}$ person = ${}^{p+q+r}{C}_p$
$2^{nd}$ person = ${}^{q+r}{C}_q$
$3^{rd}$ person = ${}^r{C}_r$
Total $=$ $\frac{(p+q+r)!}{p!(q+r)!}\frac{(q+r)!}{q!r!}\frac{r!}{r!}$

$=$ $\frac{(p+q+r)!}{p!q!r!}$
$\to$ Additionally, since $p,q,r$ are different, $1^{st}$ person could have get $q$, $2^{nd}$ r, $3^{rd}$ $p$.
$\to$ So, there are $3!$ such ways to rearrange $p,q,r$ among these persons.
So, the total number $=3!\times$ $\frac{(p+q+r)!}{p!q!r!}$