Question:

We need to count triangles formed by vertices of a $k$-sided polygon where no side of the triangle coincides with any side of the polygon. The total ways to choose 3 vertices from $k$ vertices is $\binom{k}{3}$. We must subtract triangles that share at least one side with the polygon. Triangles sharing exactly one side with the polygon: there are $k$ sides, and for each side we can choose 1 more vertex from the remaining $k-3$ vertices (excluding the 2 endpoints and adjacent vertices to avoid another side), giving $k(k-4)$ such triangles. Triangles with two consecutive sides of the polygon (using 3 consecutive vertices): there are exactly $k$ such triangles. Therefore, valid triangles = $\binom{k}{3} - k(k-4) - k = \frac{k(k-1)(k-2)}{6} - k^2 + 4k - k = \frac{k(k-1)(k-2)}{6} - k^2 + 3k = 50$. Simplifying: $\frac{k(k-1)(k-2) - 6k^2 + 18k}{6} = 50$, which gives $k(k-1)(k-2) - 6k^2 + 18k = 300$. Testing $k=10$: $10 \cdot 9 \cdot 8 - 600 + 180 = 720 - 420 = 300$. ✓