What is the number of $m$-tuples $\left(x_{1}, x_{2}, \ldots, x_{m}\right) \in\left(\mathbb{N}^{*}\right)^{m}$ satisfying $x_{1}+\ldots+x_{m}=n$?
For each $m$-tuple $\left(x_{1}, \ldots, x_{m}\right)$ of strictly positive integers such that $x_{1}+ \cdots + x_{m}=n$, we can associate the following representation.

![](https://cdn.mathpix.com/cropped/2024_05_10_0a55a0465f5204506388g-34.jpg?height=234&width=1105&top_left_y=183&top_left_x=424)

Choosing a suitable $m$-tuple is equivalent to choosing the positions of the $m-1$ separators. There are $n-1$ possible positions. Therefore, the answer is $\binom{n-1}{m-1}$.