25. Let $\dot{P}$ be a 40 -sided convex polygon. Find the number of triangles $S$ formed by the vertices of $P$ such that any two vertices of $S$ are separated by at least two other vertices of $P$.
25. Ans: 7040
For better understanding, we consider the general case when $P$ has $n$ vertices, where $n \geq 9$. We first count the number of such triangles $S$ having a particular vertex $A$. The number is $\binom{n-7}{2}$. (This can be obtained as follows. Let the triangle be $A B C$ ordered clockwise. Then $B$ has a "left" neighbour and $C$ has a "right" neighbour. The location of of $B$ and $C$ are uniquely determined by their neighbours. Besides $A, B, C$ and the two vertices to the left and two vertices to the right of $A$, the two neighbours can be chosen from the remaining $n-7$ vertices. Thus the required answer is $\frac{n}{3}\binom{n-7}{2}=7040$.