For example, placing $n$ identical balls into $m$ different boxes arbitrarily, with no limit on the number of balls in each box, how many different ways are there to do this?



Putting the above text into English while preserving the original text's line breaks and format, the translation result is as follows:
If we arrange $m$ boxes in a row and place $n$ balls into these $m$ boxes, also arranged in a row, the problem can be transformed into an occupancy problem. That is, treating $n$ balls and $m-1$ box dividers as elements, one way to arrange $n$ balls (or one way to arrange $m-1$ dividers) is one way to place $n$ balls into $m$ boxes. Clearly, there are $C_{\mathrm{n}+\mathrm{m}-1}^{\mathrm{n}}$ ways to do this.

If we let the number of balls in the $i$-th box be $x_{1}$ (where $0 \leqslant x_{1} \leqslant n$, $x_{1}$ is an integer, and $i=1,2, \cdots, m$), then the problem can be transformed into finding the number of solutions to the indeterminate equation:
$$
x_{1}+x_{2}+\cdots+x_{\mathrm{m}}=n\left(0 \leqslant x_{1} \leqslant n\right)
$$

The number of solutions to this equation is the coefficient of $x^{\mathrm{a}}$ in the expansion of the polynomial power:
$$
\begin{aligned}
(1+x & \left.+x^{2}+\cdots+x^{n}\right)^{m}=\left(\frac{1-x^{n+1}}{1-x}\right)^{-m} \\
= & \left(1-x^{s+1}\right)^{m}(1-x)^{-m}
\end{aligned}
$$

If we consider the $m$ different boxes as $m$ different elements $a_{1}, a_{2}, \cdots, a_{\mathrm{m}}$, and placing $k$ balls in the $i$-th box represents that element $a_{1}$ appears $k$ times (where $1 \leqslant i \leqslant m$ and $0 \leqslant k \leqslant n$), then the problem can be transformed into a combination problem with repetition. That is, finding the number of combinations of $n$ elements taken from $m$ different elements (with repetition allowed). This is $C_{\mathrm{n}+\mathrm{m}-1}^{a}$.

If we view each "solution" and its method as a pattern, it can be more broadly applied and extended in different directions. Consider the following two problems:
(1) Place $n$ identical balls into $m(m \leqslant i)$ different boxes, with no empty boxes. How many ways are there to do this?

This problem differs from the previous example in that it specifies no empty boxes, so it can also be solved using the above method.

If we use the occupancy problem method, note that "no empty boxes" means that the $m-1$ dividers cannot occupy two adjacent positions or the two end positions (which would result in an empty box), but can only occupy $n-1$ positions among the balls. Therefore, there are $C_{\mathrm{n}-1}^{\mathrm{m}-1}$ different ways to place the balls.

If we use the method of finding the number of solutions to an indeterminate equation, we set the number of balls in the $i$-th box to $x$. The problem then becomes finding the number of solutions to the indeterminate equation:
$$
x_{1}+x_{2}+\cdots+x_{\mathrm{m}}=n\left(1 \leqslant x_{1} \leqslant n-m+1\right)
$$

The number of solutions to this indeterminate equation is the coefficient of $x^{\mathrm{n}}$ in the polynomial power:
$$
\left(x+x^{2}+\cdots+x^{11-\mathrm{m}-1}\right)^{\mathrm{m}}
$$

(2) From the $n+m$ consecutive integers $1,2,3, \cdots, n+m$ (where $n+1 \geqslant m$), how many ways are there to choose $m$ numbers such that no two chosen numbers are adjacent?

Arrange the $n+m$ numbers in ascending (or descending) order. The chosen $m$ numbers not being adjacent means that the positions of these $m$ numbers are not adjacent. The problem can be transformed into the occupancy problem in (1). The number of ways to choose $m$ numbers such that no two are adjacent is $C_{\mathrm{n}+1}^{\mathrm{m}}$.