Number or regions formed when $n$ points on a circle are joined The maximum number $R_{n}$ of regions formed when $n$ points on a circle are joined in pairs is $\frac{1}{24}\left(n^{4}-6n^{3}+23n^{2}-18n+24\right)$.
This is a fact that I have read in several essays on the dangers of jumping to conclusions in mathematics.
In your opinion, what's the quickest (and/or nicest) way to prove this formula?
Professor Paul Zeitz explains somewhere an especially convincing way to tackle this pearl following an idea which a certain high schooler in one of his talks (lectures?) came up with; yet, I don't remember  where it was that I actually read this...
Let me thank you in advance for your insightful replies.
 A: A quick proof of the formula works using (Euler's) polyhedron theorem.
In the picture we have $\binom{n}4$ intersections inside and $n$ along the circle; this is $V=\binom{n}4+n$ vertices.
The degree of each boundary vertex is $n+1$ ($2$ arcs and $n-1$ diagonals); each interior intersection point has degree $4$. So, the sum of degrees in the graph is $n\cdot(n+1)+\binom{n}4\cdot 4$, and therefore the number of edges is $E=\frac12\left( n\cdot(n+1)+\binom{n}4\cdot 4 \right) = \frac{n(n+1)}2 +2\binom{n}4$.
If there are $R$ regions inside the circle and one outside then the polyhedron theorem gives
$(R+1) + V = E +2$, so
$$ R = 1+E-V = 1 +\left(\frac{n(n+1)}2+2\binom{n}4\right) - \left(\binom{n}4+n\right) = 1+\binom{n}2+\binom{n}4.
$$
A: The presence of the simple formula 
$$
\binom{n}4+\binom{n}2+1
$$
involving terms with a combinatorial interpretation suggests the existence of a more direct bijective proof, and indeed there is one. Assume that no two of the $n$ points have the same $y$-coordinate, and that none of the points are on the exact bottom or top of the circle; if this is not true, rotating the circle a tiny bit makes it so. 


*

*The $\binom{n}4$ counts subsets of the $n$ points of size $4$. Each of these determine a pair of intersecting chords, which in turn corresponds to one of the regions. Namely, if you look at the halves of the each of the two chords which extend upward from the intersection point, then the region is the one which borders both of these. 

*The $\binom{n}2$ term counts pairs of points. Each pair determines a chord, which corresponds to a region as follows. Looking at the lower endpoint of the chord. If that endpoint is on the right half of the circle, then the region is the one touching the lower endpoint and which is directly clockwise of the chord. If that endpoint on the left, then the process is similar, but you instead look anti-clockwise.

*At this point, all but one of the regions are accounted for. Namely, the bottommost region which contains the "south pole" of the circle is leftover, so the formula is made complete by the $+1$ term.
This process is indeed reversible; each region in the circle, except the bottommost one, determines either a quadruple or pair of points. Specifically, given a region, look at its bottommost point. 


*

*If this point is in the interior of the circle, then it corresponds to a pair of chords, which corresponds to a set of four points. 

*If this point is one of the $n$ points the border, and it is on the right side, then the that region determines a chord by looking at the anti-clockwise border. If on the left side, instead look at the clockwise border.
A: Note that the number of line segments created by connecting $n$ points is $\binom{n}{2}$ (one for every pair of points) and the number of intersection points is $\binom{n}{4}$ (one for every four points).  With no points, we have one region. Each time we add a line segment with $k$ intersections, we add $k+1$ regions (by splitting $n+1$ regions); think of this as one region for each line segment and one region for each intersection.  Thus, the number of regions is
$$
\binom{n}{4}+\binom{n}{2}+\binom{n}{0}\tag{1}
$$

Since $\binom{n}{k}=0$ for $n\gt k$, we can write
$$
2^n=(1+1)^n=\sum_{k=0}^\infty\binom{n}{k}\tag{2}
$$
and likewise
$$
0^n=(1-1)^n=\sum_{k=0}^\infty(-1)^k\binom{n}{k}\tag{3}
$$
Adding $(2)$ and $(3)$ and dividing by $2$ yields
$$
2^{n-1}=\sum_{k=0}^\infty\binom{n}{2k}\tag{4}
$$
Therefore, for $1\le n\le5$, we have
$$
\binom{n}{4}+\binom{n}{2}+\binom{n}{0}=2^{n-1}\tag{5}
$$
However, this pattern breaks down for $n\ge6$; this is because the $k=3$ term of $(4)$ is no longer $0$.
Thus, using the right side of $(5)$ to answer the question about dividing the interior of a circle works only up to a point. Furthermore, since the correct answer for $6$ points is $31$, students tend to recount the number of regions in their drawing several times.
