2. Given a regular 20-gon $M$. Find the number of quadruples of vertices of this 20-gon that are the vertices of convex quadrilaterals, which have at least one pair of parallel sides.
Answer: 765.

Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{20}$ in a circle. Each quadrilateral with a pair of parallel sides is determined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{20}$.

Consider a chord connecting two adjacent vertices of the polygon, for example, $K_{6} K_{7}$. There are 9 more chords parallel to it ( $K_{5} K_{8}$, etc.), i.e., we get a set of 10 parallel chords. From these, we can form $C_{10}^{2}=45$ pairs of parallel segments.

We will get similar sets if we consider all chords parallel to $K_{1} K_{2}, \ldots, K_{10} K_{11}$, a total of 10 such sets.

Now, let's take a chord connecting vertices that are one apart from each other, for example, $K_{6} K_{8}$. There are 8 more chords parallel to it ( $K_{5} K_{9}$, etc.), i.e., we get a set of 9 parallel chords. From these, we can form $C_{9}^{2}=36$ pairs of parallel segments. There are also 10 such sets.

In total, we get $10 \cdot 45 + 10 \cdot 36 = 810$ quadrilaterals. With this method of counting, rectangles have been counted twice. Note that both diagonals of an inscribed rectangle in a circle are diameters, and there are 10 diameters with vertices at the given points, thus we get $C_{10}^{2}=45$ rectangles. Therefore, we have $810 - 45 = 765$ quadrilaterals with at least one pair of parallel sides.