Parallelograms in the Intersections

Claim: Number of parallelograms formed $C (m,n)=^{m}C_{2}×^{n}C_{2}$

Proof: Choosing $2$ lines from $m$ parallel lines and $2$ lines from $n$ parallel lines would give an unique parallelogram(parallelogram is a quadrilateral formed by 2 pairs of parallel lines). Conversely, every parallelogram can be obtained by choosing these pairs of parallel lines. This establishes the bijection between "parallelogram" and "choose 2 lines from m, choose 2 lines from n". $_\square$

Thus, $\sum C (101, n) = ^{101} C_2 \times ^n C _2$ . Factoring $^{101}C_{2}$ out of summation, the other term is

$^{2}C_{2}+^{3}C_{2}+...+^{101}C_{2}$ . Write $^{2}C_{2}$ as $^{3}C_{3}$ as both are $1$ . Now using the identity $^{n}C_{r-1}+^{n}C_{r}=^{n+1}C_{r}$ ,

we get

$^{3}C_{3}+^{3}C_{2}+^{4}C_{2}+....+^{101}C_{2}=^{4}C_{3}+^{4}C_{2}+....+^{101}C_{2}=....=^{101}C_{3}+^{101}C_{2}=^{102}C_{3}.$

So our final answer is $^{101}C_{2}×^{102}C_{3}=\boxed {867085000}$