21. Let $P$ be a 30 -sided polygon inscribed in a circle. Find the number of triangles whose vertices are the vertices of $P$ such that any two vertices of each triangle are separated by at least three other vertices of $P$.
21. Ans: 1900

Let $A$ be a vertex of $P$. First we shall count the number of such triangles having $A$ as a vertex. After taking away $A$ and 3 consecutive vertices of $P$ on each side of $A$, we are left with 23 vertices from which we can choose two vertices in such a way that, together with $A$, a desired triangle can be formed. There are $\binom{2+(23-3-2)}{2}=\binom{20}{2}$ ways to do so. Hence there are $30\binom{20}{2} \div 3=1900$ such triangles.