Each side of triangle $ABC$ is divided into 8 equal segments. How many different triangles exist with vertices at the division points (points $A, B, C$ cannot be vertices of the triangles), such that no side of the triangle is parallel to any side of triangle $ABC$?
Let $N_{\mathrm{a}}$ be the number of triangles for which one of the sides is parallel to side $B C$ of the original triangle. Similarly, we define the numbers $N_{\mathrm{b}}, N_{\mathrm{c}}, N_{\mathrm{a}, \mathrm{b}}, N_{\mathrm{b}, \mathrm{c}}, N_{\mathrm{a}, \mathrm{c}}$ and $N_{\mathrm{a}, \mathrm{b}, \mathrm{c}}$. Let $N$ denote the total number of triangles. Then $N=7^{3}, N_{\mathrm{a}}=N_{\mathrm{b}}=N_{\mathrm{c}}=7^{2}, N_{\mathrm{a}, \mathrm{b}}=N_{\mathrm{b}, \mathrm{c}}=N_{\mathrm{a}, \mathrm{c}}=7, N_{\mathrm{a}, \mathrm{b}, \mathrm{c}}=1$. The desired number is found using the inclusion-exclusion principle:

$$
7^{3}-3 \cdot 7^{2}+3 \cdot 7-1=6^{3}
$$