(Dur) Consider a triangular grid obtained by dividing an equilateral triangle of side $n$ into $n^{2}$ equilateral triangles of side 1. Determine the number of parallelograms that appear on this grid.

## Pot-pourri
Note that a parallelogram within the triangle can have one of three orientations, given by the side of the triangle that is not parallel to the sides of the parallelogram. There are the same number of parallelograms of each orientation, so it suffices to count those with no side parallel to the base of the triangle (see figure below). Such a parallelogram is defined by four lines that come from four points on the base of the triangle. Any choice of four points from the $n+1$ points on the base of the triangle then defines a unique parallelogram (as shown in the figure). However, we must be careful because we have not counted the parallelograms that touch the base of the triangle, which correspond to the case where the second and third points are coincident. This case corresponds to the choice of three points on the base of the triangle, and the number of parallelograms in the triangle is (don't forget to multiply by 3 to account for the orientation):

$$
3\left(\binom{n+1}{4}+\binom{n+1}{3}\right)=3\binom{n+2}{4}
$$

by Pascal's formula.

![](https://cdn.mathpix.com/cropped/2024_05_10_10c029b6c4865f8b8ca3g-161.jpg?height=560&width=625&top_left_y=254&top_left_x=721)