Problem 10-3. Grisha drew a convex 100-gon on the plane and drew all its diagonals, and, wonder of wonders, no more than two segments intersected at any point other than the vertices of the 100-gon. In how many ways can Grisha highlight some of the lines on the drawing with a marker to form a triangle (not necessarily consisting of whole diagonals and, perhaps, containing unhighlighted lines inside it)?

$$
\text { Answer: }\binom{100}{3}+4\binom{100}{4}+5\binom{100}{5}+\binom{100}{6}
$$
Solution. Let's find the ends of the diagonals outlined by Grisha. Each diagonal has two ends, but some ends can coincide, but no more than two in one point. Thus, we will get from 3 to 6 vertices of the hundred-gon. We will find the corresponding number of triangles.

For three vertices, the number of triangles is equal to the number of ways to choose three different vertices $\binom{100}{3}$.

For four vertices, a pair of adjacent ones forms a side of the triangle (the other two do not coincide, but only contain them). Thus, the number is $4\binom{100}{4}$.

For five vertices, we need to choose five vertices and among them one that is a vertex of the triangle. Thus, the number is $5\binom{100}{5}$.

For a hexagon, the way is unique after choosing six vertices, that is, $\binom{100}{6}$.

In total, we get $\binom{100}{3}+4\binom{100}{4}+5\binom{100}{5}+\binom{100}{6}$.

| Rating | Score | Content of the criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution (including for a p-gon) |
| .+ | 15 | Minor gaps in the solution (including not proving that triangles can be matched with p-gons) |
| $+/-$ | 10 | Serious gaps in the proof |
| .- | 5 | The idea of dividing triangles into 4 types |
| - | 0 | The solution is completely incorrect/only the answer |