The number of ways in which $5$ different balls can be placed in $3$ identical boxes such that no box remains empty is

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
$\frac{5!}{1!\cdot 2!\cdot 3!}+\frac{5!}{(2!{)}^{2}1!\cdot 2!}=25$