Consider parallelograms parallel to sides AB and AC.
We extend the n row of the triangle to the n+1 rows.
For any 4 distinct points, P1, P2, P3, P4 from left to
right, we draw slant lines like "/" at P1 and P2, and
"\" slant lines at P3 and P4.  They will bound one
single parallelogram.  Also for any parallelogram we
can extend the sides so it lands on the same four
distinct points on the n+1 row.  So there is a
bijection between the number of ways to choose 4
distinct points on n+1 row and number of
parallelograms in an equilateral triangle of side
length n.  there are n+2 points in n+1 row so there
are C(n+2,4) ways of choosing 4 points.  By complete
symmetry the total number of parallelograms is 3C(n+2,4)