【Example 4】There are $n$ types of books, with each type having no less than 10 copies. If you randomly select 5 books to give away, how many different ways can this be done?
Consider the problem of placing 5 indistinguishable balls into $n$ distinct boxes. If the number of balls placed in a certain box represents the number of corresponding books taken out, then the original book-taking problem is transformed into a ball-into-box problem. Since the capacity of each box (at least 10 books) is greater than the number of balls, it is equivalent to a placement problem with unlimited capacity. Therefore, the total number of different ways to do this is
$$
C_{n+5-1}^{5}=C_{n+4}^{5}=\frac{1}{5!}(n+4)(n+3)(n+2)(n+1) n \text{ (ways)}
$$

In the previous section, we discussed the "empty box problem" for distinguishable balls. Now, we need to discuss this problem for indistinguishable balls. Here, the boxes are assumed to be distinct, each with unlimited capacity, and the number of balls $k \geqslant$ the number of boxes $n$. We now calculate the number of scenarios where no empty boxes appear.

We can still imagine using $n-1$ dividers to separate $k$ balls arranged in a row into $n$ segments, but no two dividers can be adjacent, otherwise an empty box would appear. Similarly, dividers cannot be placed at the ends. Therefore, it is equivalent to selecting $n-1$ positions from the $k-1$ gaps between the $k$ balls (Figure 14), which is a combination problem. Thus, there are $C_{k-1}^{n-1}$ different scenarios.