We are given a regular polygon with 200 sides. How many triangles can be formed whose three vertices are vertices of this polygon?
Answer: 1313400 We choose a first vertex: there are 200 choices. We choose a second vertex: there are 199 possibilities. Then, there are 198 possibilities left for the third vertex. This makes a total of $200 \times 199 \times 198$ triangles. However, each triangle has been counted 6 times because triangle $A B C$ is the same as triangles $A C B, B A C, B C A, C A B, C B A$. The answer is therefore $200 \times 199 \times 198 / 6$.