, 2011

The parallelograms can be naturally partitioned into three sets, based on the orientation of their sides in the grid. By symmetry there will be the same number of parallelograms of each orientation, so we can restrict ourselves to the 'diamond' case (as in the diagram).

Extend the triangular grid by one. The sides of the parallelogram can then be extended until they cut the new bottom row of the grid. It is easy to see that every parallelogram will correspond to a different set of four intersection points (marked $a$, $b$, $c$ and $d$ for the case of the shaded parallelogram), and that every set of four points will correspond to a parallelogram. This bijection makes it easy to count – there are $n+2$ points on the bottom row, so there are $\binom{n+2}{4}$ ways of making a parallelogram. Taking into account the orientations, the final answer is $3\binom{n+2}{4}$.