Five racquets need to be placed in three boxes. Each box can hold all the five racquets. In how many ways can the racquets be placed in the boxes so that no box can be empty if all racquets are different but all boxes are identical?

Let the 5 racquets be a, b, c, d, e

3 Boxes are Identical

Since the boxes are identical, 3 in First box, 1 in Second and 1 in third is not different from 1 in first, 3 in second and 1 in third.

$\begin{array}{|c|c|c|} \hline \text{Box 1} & \text{Box 2} & \text{Box 3} \\ \hline 3 \text{ racquets} & 1 \text{ racquet} & 1 \text{ racquet} \\ 5C_3 \text{ ways} & 2C_1 \text{ ways} & 1C_1 \text{ way} \\ \hline 2 \text{ racquets} & 2 \text{ racquets} & 1 \text{ racquet} \\ 5C_2 \text{ ways} & 3C_2 \text{ ways} & 1C_1 \text{ way} \\ \hline \end{array}$

But there is a catch in the above table.

Imagine this process, after selecting 3 out of a, b, c, d, e

Let's say we select d, e, c. Then a and b have to be put in two identical boxes.

This can be done only in one way.

In second scenario let's assume a and b in First box and c and d in second box and e in third box. But the possibility will be same when c and d in First box and a and b in second box.

So, it is $\frac{^5C_2 \times ^3C_2}{2} = 15$ ways, In first scenario $^5C_3 = 10$ ways

Total $=25$ ways.