Question

No. of ways of distributing $(2+3+5+8)$ different things to $4$ persons so that one person gets $2$ things, 2nd person get $3$ things, 3rd person get $5$ things and 4th person get $8$ things is

• A. $\frac{18!}{4!.2!.8!}$
• B. $\frac{18!}{2!.3!.5!.8!}\times 4!$
• C. $\frac{18!}{4!.2!.3!.5!.8!}$
• D. $2!.3!.5!.8!$

Answer 1

Answer: (B)

$\frac{18!}{2!.3!.5!.8!}\times 4!$

$\to$ No. of ways of distributing $(p+q+r+s)$ different things to $4$ persons so that one gets $p$, second gets $q$, third gets $r$ and fourth gets $s$ things is given by
$=\frac{(p+q+r+s)!}{p!q!r!s!}\times 4!$

Here, we need to distribute $(2+3+5+8)$ things to $4$ people

So, $p=2,q=3,r=5$ and $s=8$
$\therefore$ Total no. of ways $=\frac{(2+3+5+8)!}{2!3!5!8!}\times 4!$

$=\frac{\left(18\right)!}{2!3!5!8!}\times 4!$
Hence, option B is correct.