Question

Let $m$ denote the number of ways in which four different balls of green colour and four different balls of red colour can be distributed equally among $4$ persons if each person has balls of the same colour and $n$ be the corresponding figure when all the four persons have balls of different colour. Then the value of $m+n$ is

• A. 792.00
• B. 792.0
• C. 792

Answer 1

Answer: (A), (B), (C)

$4$ green balls $(G_1,G_2,G_3,G_4)$ and $4$ red balls $(R_1,R_2,R_3,R_4)$ are to be distributed among $4$ persons.

$4$ green balls can be divided into two equal
groups in $\frac{4!}{2!2!2!}=\frac{4!}{8}$ ways
Similarly, $4$ red balls in $\frac{4!}{2!2!2!}=\frac{4!}{8}$ ways.
Number of four equal groups $=\left(\frac{4!}{8}\right)\left(\frac{4!}{8}\right)$
Now, we have $4$ equal groups. Each equal group has $2$ balls of the same colour. They can be distributed in $4$ persons in $4!$ ways.
$\therefore m=\left(\frac{4!}{8}\right)\left(\frac{4!}{8}\right)(4!)=216$

Now, $n={\left({}^4{C}_1\right)}^2{\left({}^3{C}_1\right)}^2{\left({}^2{C}_1\right)}^2{\left({}^1{C}_1\right)}^2=(24{)}^2$
$=576$
$[{\left({}^4{C}_1\right)}^2\to 1$ out of $4$ green and $1$ out of $4$ red can be taken in $\left({}^4{C}_1\right)\left({}^4{C}_1\right)={\left({}^4{C}_1\right)}^2$
ways.]

Hence, $m+n=216+576=792$