If there are N points on a circle, and you draw a chord between each of them, how many regions is the circle subdivided into?

I'm not quite sure where to begin. I know that there are at least n section for n points on a chord because if you connect each point on the circle with its adjacent point then that forms n sections. Whatever goes on in the middle though seems really arbitrary. Is there any good way of approaching this problem?


  • $\begingroup$ Do you suppose that the instruction is to draw a chord between each pair of them (points on a circle)? Or that you merely find a minimal set of chords such that every pair of points is separated by (at least) one chord? $\endgroup$
    – hardmath
    Nov 26, 2016 at 2:53
  • $\begingroup$ Euler's formula could kill it $\endgroup$
    – x100c
    Sep 23, 2021 at 14:09

2 Answers 2


Let's assume that no three lines intersect at the same point.

Consider what happens each time you add a line between two points. If it crosses $k$ other lines, it splits $k+1$ regions into two, thereby adding $k+1$ regions. Therefore, when a line is added, it can be considered to be adding $1$ region for the line and $1$ region for each point of intersection:

$\hspace{4.5cm}$chords $$ \text{Regions Split by a Line}\atop\text{Each Region is Paired with a Line or Point of Intersection} $$ Thus, we just need to count $1$ for the original region inside the circle, and add the number of lines, $\binom{N}{2}$, and the number of intersections, $\binom{N}{4}$.

Thus, there are $\binom{N}{4}+\binom{N}{2}+\binom{N}{0}$ regions.

Counting Lines and their Intersections

Note that given $N$ points on the circle, each pair of points yields a line, thus the number of lines connecting the points is $\binom{N}{2}$. For each set of $4$ points, there is a unique intersection of two lines, an "X in a quadrilateral" arrangement. Thus, the number of interior line intersections is $\binom{N}{4}$.


This is often given as an example of what happens if you attempt to guess a sequence from the first few terms since this sequence starts with $$ 1,2,4,8,16,\dots $$ but the next term is $31$.


The reason for the initial similarity of $\binom{n}{4\vphantom{0}}{+}\binom{n}{2\vphantom{0}}{+}\binom{n}{0}$ to the geometric progression is that $$ \sum_{k=0}^\infty\binom{n}{k}=(1+1)^n=2^n\tag{1} $$ and $$ \sum_{k=0}^\infty(-1)^k\binom{n}{k}=(1-1)^n=0^n\tag{2} $$ For $n>0$, adding $(1)$ and $(2)$ and dividing by $2$ gives $$ \sum_{k=0}^\infty\binom{n}{2k}=2^{n-1}\tag{3} $$ $\binom{n}{4\vphantom{0}}{+}\binom{n}{2\vphantom{0}}{+}\binom{n}{0}$ is the first three terms of $(3)$. The first time a non-zero term is left out of $(3)$ is when $n=6$, and that is when $\binom{n}{4\vphantom{0}}{+}\binom{n}{2\vphantom{0}}{+}\binom{n}{0}=31$.

  • $\begingroup$ Could you say a bit more about where exactly the $\binom{n}{2}$ and the $\binom{n}{4}$ come from, please? $\endgroup$ May 19, 2012 at 23:25
  • $\begingroup$ @Peter: I've added a small section on that. Let me know if you would like more. $\endgroup$
    – robjohn
    May 20, 2012 at 0:22



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