Question

In how many ways $5$ different balls can be distributed into $3$ boxes so that no box remains empty?

Answer 1

Since each box should have atleast one ball,the remaining 2 can be divided in the following ways- $(3,1,1)$ or $(2,2,1)$.

So total number of ways of distributing among the 3 boxes would be $[\frac{5!}{3!}\times 2!]\times 3!+\left[\frac{5!}{2!2!2!}\right]\times 3!$

$=60+90=150.$