Balls and Boxes gone crazy. Part-III

Since each box must have at least one ball placed in it, the only possible "ball counts" are (i) two boxes each with one ball apiece and the third with three balls, and (ii) two boxes with two balls apiece and the third with one ball.

For case (i), we can choose two of the five balls in $\binom{5}{2} = 10$ ways to be placed alone in separate boxes, with the remaining three balls going in the remaining box.

For case (ii), we can choose one of the five balls to be placed alone in one box in $5$ ways. We can then "pair off" the remaining four balls in $3$ ways to be placed pairwise in the remaining two boxes. This then gives us $5*3 = 15$ ways this case can play out.

Thus there are a total of $10 + 15 = \boxed{25}$ ways we can place the five distinct balls in the three identical boxes.

In other words, we have to divide 5 distinct balls to 3 identical boxes such that no box is remained empty. Possible configuration for boxes are: permutations of (2,2,1) and (3,1,1) . So, total number of ways of doing (2,2,1) divisions is (5!/2!2!1!)/2!. Now, since boxes are identical, hence total number of ways of doing divisions = 15

Similarly, total number of ways of doing (3,1,1) divisions is (5!/3!1!1!)/2!. Now, since boxes are identical, hence total number of ways of doing divisions = 10

So, by Fundamental principle of counting, total number of ways = 15+10 = 25

Referring to the case when the three boxes are different ( Balls and Boxes gone crazy. Part-I ), total ways we could place $5$ different balls in $3$ different boxes (such that no box is empty) $= 150$ .

Now, with the boxes being identical, each way to place the balls in three boxes (such that no box is empty) has been counted $3!$ times.

Hence total ways to place balls in three boxes such that no box is empty and the boxes are identical $= 150/3! = 25$ .