1. Each of the six houses on one side of the street is connected by cable lines to each of the eight houses on the opposite side. How many pairwise intersections do the shadows of these cables form on the surface of the street, if no three of them intersect at the same point? Assume that the light causing these shadows falls vertically downward.
# Solution

Let's take an arbitrary pair of houses on one side of the street and an arbitrary pair on the other. They are the vertices of a convex quadrilateral (since two sides of the quadrilateral, coming from each chosen pair, lie on one side of the line, i.e., the angles do not exceed $180^{\circ}$), so its diagonals intersect.

Each pairwise intersection of shadows (cables) is a point of intersection of the diagonals of such a quadrilateral. Thus, it remains to find their number, which is equal to the product of the ways to choose an ordered pair of houses on each side of the street.

Answer: $C_{6}^{2} \cdot C_{8}^{2}=420$.