[Math] no of intersection points.!

combinatorics

Prove that if each of $m$ points in one straight line be joined to each of $n$ in a parallel straight line by straight line segments terminating at the points, then excluding the given points, the line segments will intersect $\frac14mn(m-1)(n-1)$ times. Note: The count is of intersections, not of points of intersection. E.g., in the picture below for the case $m=n=3$, each red dot represents one intersection, but the large black dot in the centre represents three intersections, one for each pair of segments intersecting at that point. (I did not bother to include the original straight lines on which the points lie; they’re pretty obvious and just clutter up the picture.)

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Best Answer

There's a problem if we allow the lines to intersect, for example:

Problem with intersecting lines

results in no intersections.

If the two lines are parallel, we have a unique intersection for any pair of $2$-element subsets: one from each line. This is illustrated below:

Parallel lines illustration

Conversely, any such intersection occurs between such pairs. The number of such pairs is $$\binom{m}{2}\binom{n}{2},$$ as required.

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