Question

In how many ways $8$ balls can be distributed among $3$ children so that one child gets $4$ balls and two children get $2$ ball each?

• A. $210$
• B. $240$
• C. $420$
• D. $1260$

Answer 1

Answer: (D)

$1260$

Let there be $3$ children $A,B$ and $C.$

$Case$ $1:$ $A$ gets $4,B$ and $C$ gets $2$ each.

Total no. of ways this can be done $={8C}_4\times {4C}_2\times {2C}_2$

$A,B,C$ can be arranged in $\frac{3!}{2!},$

Here that sequence in which each of them receive the ball is important .

$\therefore \text{Total ways}=({8C}_4\times {4C}_2\times {2C}_2)\frac{3!}{2!}=1260$ ways.