How many can you make?

The first set of 10 parallel lines are parallel to AB and thus CD. Now we have to choose 2 lines among these 12 lines which can be found by ${\phantom{0}}^{12}C_{2}$ (10 lines + 2 sides). Similarly, we have to choose two lines from the second set which are parallel to BC and thua DA. Thus too can be found by ${\phantom{0}}^{12}C_{2}$ . So, the total number of parallelograms that can be formed $= {\phantom{0}}^{12}C_{2} × {\phantom{0}}^{12}C_{2}\\ = \dfrac{12!}{2! × 10!} × \dfrac{12!}{2! × 10!}\\ = 66 × 66\\ = \boxed{4356}$ .