2. Given a regular 16-gon $M$. Find the number of quadruples of vertices of this 16-gon that are the vertices of convex quadrilaterals, which have at least one pair of parallel sides.
Answer: 364.

Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{16}$ in a circle. Each quadrilateral with a pair of parallel sides is determined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{16}$.

Consider a chord connecting two adjacent vertices of the polygon, for example, $K_{6} K_{7}$. There are 7 more chords parallel to it ( $K_{5} K_{8}$, etc.), i.e., we get a set of 8 parallel chords. From these, we can form $C_{8}^{2}=28$ pairs of parallel segments.

We will get similar sets if we consider all chords parallel to $K_{1} K_{2}, \ldots, K_{8} K_{9}$ - a total of 8 such sets.

Now, let's take a chord connecting vertices that are one apart, for example, $K_{6} K_{8}$. There are 6 more chords parallel to it ( $K_{5} K_{9}$, etc.), i.e., we get a set of 7 parallel chords. From these, we can form $C_{7}^{2}=21$ pairs of parallel segments. There are also 8 such sets.

In the end, we get $8 \cdot 28 + 8 \cdot 21 = 392$ quadrilaterals. With this method of counting, rectangles were counted twice. Note that both diagonals of an inscribed rectangle in a circle are diameters, and there are 8 diameters with vertices at the given points, thus we get $C_{8}^{2}=28$ rectangles. Therefore, we have $392 - 28 = 364$ quadrilaterals with at least one pair of parallel sides.