Moscow Seven Sisters

Question

Fix $n$ points in the plane in generic position, i.e. no three of them on the same line, etc. The number of lines joining two of them is ${n \choose 2}$. The number of regions in which $\ell$ lines cut the plane is $\leq \ell (\ell+1)/2+1$. By moving around in the plane, we observe the $n$ points in different circular orderings. Since there are $(n-1)!$ such orderings, it is clear that for $n \geq 7$ we cannot observe the points in all distinct orderings, while for $n \leq 4$ this is possible. What happens for $n=5$ and 6?

Answer 1

Here is just a small correction, the maximum number $F_n$ of possible regions created by straight lines passing through a given set of $n$ points is $$F_n:=\begin{cases}1&\text{if }n\in\{0,1\}\,,\\3\binom{n}{4}+3\binom{n}{2}-n+1&\text{if }n=2,3,4,\ldots\end{cases}\,.$$ It is easy to show that $$F_n<(n-1)!\text{ for }n=6,7,8,\ldots\,.$$

Thus, it remains to verify whether all cyclic permutations of the $n$ points can be seen on the plane when $n=5$. After some trials, I believe that the answer for $n=5$ is that there does not exist such a point configuration. However, I have no proof.