Question

Five balls are to be placed in three boxes. Each box can hold all the five balls so that no box remains empty.

If balls are different but boxes are identical then number of ways is

Answer 1

If no box remains empty, then we can have $(1,1,3)$ or $(1,2,2)$ distribution pattern.
When balls are different and boxes are identical, number of distributions is equal to number of divisions in $(1,1,3)$ or $(1,2,2)$ ways.
Hence, total number of ways is
$\frac{5!}{1!\cdot 2!\cdot 3!}+\frac{5!}{(2!{)}^{2}1!\cdot 2!}=25$