Question

$5$ balls are to be placed in $3$ boxes. Each box can hold all the $5$ balls. Number of ways in which the balls can be placed so that no box remains empty, if :

Answer 1

$A.$
Each box (say $B_1,B_2,B_3$) will have at least one ball.

Now the ways for placing other $2$ identical balls in $3$ different boxes are

$\frac{(2+3-1)!}{2!(3-1)!}=6(\therefore \frac{(n+r-1)!}{n!(r-1)!})$

$B.$
Case $1$ : $5$ balls can be divided in $3$ groups having $2$ balls each in $2$ boxes and $1$ ball for in third box $(2,2,1)$

ways : $\frac{5!}{(1!)(2!{)}^2\times 2!}=15$

Case $2$ : Division can also be $3$ in one box and $1$ each in remaining $2$ boxes $(3,1,1)$

ways : $\frac{5!}{(3!)(1!{)}^2\times 2!}=10$

Hence total ways $=10+15=25$

$C.$
Only $2$ arrangements are possible.

1. $2$ balls each in $2$ boxes & remaining ball in other box $(2,2,1)$

2. $3$ balls in $1$ box and 1 ball each in other boxes $(3,1,1)$

$D.$
Same as part $A$