Question

Number of ways in which 9 different toys can be distributed among 4 children belonging to different age groups insuch a way that distribution among the 3 elder children is even and the youngest one is to receive one toy more is

• A. $\frac{(5!{)}^2}{8}$
• B. $\frac{9!}{8}$
• C. $\frac{9!}{3!(2!{)}^3}$
• D. None of these

Answer 1

The correct option is C $\frac{9!}{3!(2!{)}^3}$

The distribution should be $2,2,2$ and $3$ to the youngest.

Now $3$ toys for the youngest van be selected in ${}^9{C}_3$ ways, remaining $6$ toys can be divided into three equal groups in $\frac{6!}{{(2!)}^{3}3!}$ ways

and can be distributed in $3!$ ways.

Thus the required number of ways ${=}^9{C}_3\frac{6!}{{(2!)}^3}=\frac{9!}{3!{(2!)}^3}$