With 5 equal digits, I can write 100 as follows: $100=111-11$, or $100=3 \times 33+\frac{3}{3}$, or $(5+5+5+5) \times 5$. Based on these, let's write 100 with 9 different digits.
The task has multiple solutions. For example:

$$
\begin{gathered}
1+2+3+4+5+6+7+8 \cdot 9=100 \\
1+2+3+47+5 \cdot 6+8+9=100 \\
1+2+3+4+5-6-7+8+92=100
\end{gathered}
$$

Number of solutions: 38.