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I've been thinking about the classic problem where you have to determine the maximum number of regions created when you place $n$ points around the circumference of a circle and connect every pair of points with a line segment. There is a classic argument using Euler's formula which shows that the maximum number of regions is $$ 1 + {n \choose 2} + {n \choose 4} $$

Along the way, it is argued that there well be $n\choose 2$ line segments. This is clear enough. It is also argued that there are $n \choose 4$ points of intersection between the segments, as any intersection is uniquely specified by the $2$ pairs of endpoints of the segments intersecting at that point.

It seems like we could just as well have uniquely specified any intersection point by the $2$ segments that intersect at that point. In other words, any choice of $2$ of the $n \choose 2$ line segments should also uniquely specify a point of intersection. Thus, it seems that ${{n \choose 2} \choose {2}}$ should also give the number of points of intersection between the segments.

But, of course $$ {n \choose 4} \neq {{n \choose 2}\choose 2} $$ so there must be a flaw in my reasoning. Can someone help me to identify it?

Charles Hudgins
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    Take $n=4$ with points A,B,C,D around the circle in that order. Lines AB and CD don't intersect inside the circle. Etc. – Ned Oct 10 '19 at 01:23
  • There's ambiguity and missing information in your post which makes it hard to answer your question. When you refer to "lines in the graph", do you mean actual lines in the sense of Euclidean geometry which extend infinitely in both directions? Or, instead are you referring to line segments, as indicated when you refer to "the 2 pairs of endpoints of the lines"? Also, it is possible for lines to be parallel in this context, and hence not intersect at all? Is it possible for line segments to not intersect even when not parallel? – Lee Mosher Oct 10 '19 at 01:23
  • I'll expand the introduction a bit to make it more clear. – Charles Hudgins Oct 10 '19 at 01:24
  • Having chosen four of the points, $A,B,C,D$, which lines intersecting are we talking about? The one where $AB$ intersects $CD$? The one where $AC$ intersects $BD$? Or the one where $AD$ intersects $BC$? I would think the error is in the $\binom{n}{4}$ side of things, either in not specifying that we are talking only about intersections within the circle, or otherwise a miscalculation. – JMoravitz Oct 10 '19 at 01:31
  • I think the point is that it's a complete graph, so exactly one of those choices should yield an intersection. – Charles Hudgins Oct 10 '19 at 01:34

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If you look at the pairwise intersections of the $\binom n2$ lines of the form $A_j A_k$ (where $A_1,\ldots, A_n$ are the chosen points on the circle) then many of them are already just the $A_i$, since some pairs of lines look like $A_1A_2$ and $A_1A_3$. I reckon there are $n\binom{n-1}2$ pairs like this. But even when we have pairs of lines $A_i A_j$ and $A_k A_l$ with all the points distinct, then some of those pairs will intersect outside the circle. Each quartet of points will determine three pairs of lines, and only one of those pairs will intersect within the circle. Therefore the real identity is $$\binom{\binom n2}2=n\binom{n-1}2+3\binom n4.$$

Angina Seng
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  • Makes perfect sense. For reference, thinking carefully about this question is what caused me to ask this one. If you have the time, you might want to give the argument presented there a spot check to see if it actually goes through. – Charles Hudgins Oct 10 '19 at 02:13