Let $l_1$ and $l_2$ two parallel lines. We choose $8$ points on $l_1$ and $7$ on $l_2$. Then all the market points on $l_1$ are joined with the market points on $l_2$. How many points are obtained as an intersection (that are not on $l_1$ nor on $l_2$) of the traced segments?
I´ve solved the problem by doing all the lines and I got $588$ intersections, so, I´m looking for a solution without making all of them. Any hint?
If I take the first point on $l_1$ and then traced all the lines with the respective point on $l_2$ I got $0$ intersections. Then if I take the second point and joined to the first on $l_2$ I have $6$ intersections, I noticed that the next will add $5$... So, the first lines are not crossed. Then from the second point, there are $6+5+4+3+2+1$ from the third point, $12+10+8+6+4+2$, from the fourth $18+15+12+9+6+3$, from the fifth $24+20+16+12+8+4$, from the sixth $30+25+20+15+10+5$, from the seventh $36+39+24+18+12+6$, finally from the eighth $42+35+28+21+14+7$. Adding up all the numbers I get $588$.
Pick two marked points on $l_1$ and two on $l_2$. There is exactly one way to draw two segments between those points that creates an intersection; since all possible segments are drawn in the original question, the number of ways to pick the pairs of points on $l_1$ and $l_2$ is the number of intersections.
On $l_1$ a pair of points can be picked in $\binom82$ ways; on $l_2$, $\binom72$ ways. The number of intersections is $\binom82\binom72=588$.