Example 4. Find the indefinite solution of the equation
$$
x+y+z=20
$$
Find the number of positive integer solutions for (1). For this, we have 20 a's and 19 spaces (denoted by $s$) arranged in one line as follows:
asasasasasasasasasasasasasasasasas asisa.

Now, choose any 2 out of the 19 s's, and remove the rest of the $s$'s, for example:
caaaasaaaaaasaaauaaaaa. This results in three segments. From left to right, the counts are 5, 6, and 9, respectively, so we have
$$
x=5, y=6, z=9 \text {. }
$$

It is easy to see that each way of choosing 2 out of the 19 s's corresponds to a solution of (4), and different choices correspond to different solutions of (4); conversely, each solution of (4) corresponds to a way of choosing 2 out of the 19 s's, and different solutions correspond to different ways of choosing. In summary, the number of positive integer solutions for (4) is equal to the number of ways to choose 2 out of 19 s's, which is
$$
C_{19}^{2}=\frac{19 \times 18}{2}=171 .
$$

To find the number of non-negative integer solutions for (4), we can use 20 white balls and 2 black balls, arranged in any order from left to right on a straight line. These 2 black balls divide the line into 3 segments. Then, count the number of white balls in each segment from left to right, denoted as $\alpha_{1}, \alpha_{2}, \alpha_{3}$, and let
$$
x=\alpha_{1}, y=\alpha_{2}, \quad z=\alpha_{3},
$$
The number of non-negative integer solutions for (4) is equal to the number of arrangements of these 22 balls. However, note that among these 22 balls, 20 white balls are identical, and 2 black balls are identical, so the total number of arrangements is
$$
\frac{22!}{20!2!}=C_{22}^{2}=231 .
$$
$$
x_{1}+x_{2}+\cdots+x_{m}=n \quad(n \geqslant m)
$$
The number of solutions is $C_{n+m-1}^{m-1}$.