Question

Five balls of different colors are to be placed in 3 boxes of different size. Each box can hold all five. In how many different ways can we place the balls so that no box remains empty?

Answer 1

As no box should remain empty, boxes can have balls in the following numbers:
Possibilities $1,1,3$ or $1,2,2$
Division ways for tree (i) is $\frac{5!}{(1!{)}^{2}3!2!}$
Division ways for tree (ii) is $\frac{5!}{(2!{)}^{2}1!2!}$
Now, total number of ways of distribution of these groups into three boxes is
$[\frac{5!}{(1!{)}^{2}3!2!}+\frac{5!}{(2!{)}^{2}1!2!}]\times 3!=150$.