A plane contains $40$ lines, no $2$ of which are parallel. Suppose that there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Find the number of points where exactly $2$ lines intersect.
In this solution, let $\boldsymbol{n}$-line points be the points where exactly $n$ lines intersect. We wish to find the number of $2$-line points.
There are $\binom{40}{2}=780$ pairs of lines. Among them:

The $3$-line points account for $3\cdot\binom32=9$ pairs of lines.
The $4$-line points account for $4\cdot\binom42=24$ pairs of lines.
The $5$-line points account for $5\cdot\binom52=50$ pairs of lines.
The $6$-line points account for $6\cdot\binom62=90$ pairs of lines.
It follows that the $2$-line points account for $780-9-24-50-90=\boxed{607}$ pairs of lines, where each pair intersect at a single point. 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MRENTHUSIASM