Example 6: Taking three points from $2 n$ equally divided points on a circle, how many acute triangles can be formed?
Given that $\triangle A B C$ is an acute triangle, let there be $x_{1}$ points between points $A$ and $B$, $x_{2}$ points between points $B$ and $C$, and $x_{3}$ points between points $C$ and $A$. Then,
$$
x_{1}+x_{2}+x_{3}=2 n-3 \text {, }
$$

where $1 \leqslant x_{k} \leqslant n-2, k=1,2,3$.
Let $x_{i}^{\prime}=x_{i}-1$, then
$$
x_{1}^{\prime}+x_{2}^{\prime}+x_{3}^{\prime}=2 n-6\left(0 \leqslant x_{i}^{\prime} \leqslant n-3\right) \text {. }
$$

Let $x_{i}^{\prime \prime}=(n-3)-x_{i}^{\prime}$, then
$$
x_{1}^{\prime \prime}+x_{2}^{\prime \prime}+x_{3}^{\prime \prime}=n-3\left(0 \leqslant x_{i}^{\prime \prime} \leqslant n-3\right) \text {. }
$$

At this point, the equation has $\mathrm{C}_{n-1}^{2}$ solutions.
When point $A$ is fixed, there are $\mathrm{C}_{n-1}^{2}$ acute triangles.
Therefore, there are $\frac{2 n}{3} \mathrm{C}_{n-1}^{2}=\frac{n(n-1)(n-2)}{3}$ acute triangles.