How many solutions does the equation $x+y+z=100$ have in the set of positive integers?
I. solution. Let's take 100 pebbles and place them in a row next to each other. To divide them into three parts, we need to separate the row of pebbles at two points. There are 99 "gaps" between the 100 pebbles where we can do this. We choose two out of these 99 "gaps," which can be done in $\binom{99}{2}$ ways. This results in three piles: with $x ; y ; z$ pebbles.

Thus, the equation has $\binom{99}{2}=4851$ solutions in the set of positive integers.

II. solution. None of the numbers can be 99, since in that case we would still need to add at least $2 \cdot 1$, which we cannot do. We create a table:

| $x$ | $y$ | $z$ |  |
| :---: | :---: | :---: | :---: |
| 98 | 1 | 1 | 1 solution |
| 97 | 1 | 2 | 2 solutions |
| 97 | 2 | 1 |  |
| 96 | 1 | 3 | 3 solutions |
| 96 | 2 | 2 |  |
| 96 | 3 | 1 |  |
| 95 | 1 | 4 | 4 solutions |
| 95 | 2 | 3 |  |
| 95 | 3 | 2 |  |
| 95 | 4 | 1 |  |
| $\vdots$ | $\vdots$ | $\vdots$ |  |
| 2 | 1 | 97 | 97 solutions |
| 2 | 2 | 96 |  |
| $\vdots$ | $\vdots$ | $\vdots$ |  |
| 2 | 97 | 1 |  |
| 1 | 1 | 98 | 98 solutions |
| 1 | 2 | 97 |  |
| $\vdots$ | $\vdots$ | $\vdots$ |  |
| 1 | 98 | 1 |  |

After this, we need to sum the number of solutions obtained:

$$
1+2+3+\ldots+96+97+98=\frac{98 \cdot 99}{2}=4851
$$

Thus, the equation $x+y+z=100$ has 4851 solutions in the set of positive integers.

Note. In the problem, we distinguish between the "piles" according to the meaning of the variables, so, for example, the decompositions $1+98+1$ and $98+1+1$ are considered different.