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?

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    $\begingroup$ This is a very important missing piece of information, without which the question is unanswerable. Who are Moscow seven sisters and how are they related to this problem? $\endgroup$ – Batominovski Oct 19 '18 at 21:14
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    $\begingroup$ @Batominovski For $n=7$, you get seven points, which you can picture as seven buildings in the city of Moscow, Wikipedia is your friend. $\endgroup$ – jj_p Oct 19 '18 at 21:38
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    $\begingroup$ @jj_p: Requiring a reader to have to perform a web search to understand the context of a question is poor authorship. In this case, it would be most helpful for you to include your above comment as part of the question itself (not everyone reads comments), providing a specific link to whatever Wikipedia entry you believe is relevant. $\endgroup$ – Blue Oct 20 '18 at 8:56
  • $\begingroup$ @Blue That is just a curiosity, totally irrelevant to understand the math question.. $\endgroup$ – jj_p Oct 20 '18 at 17:49

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.

  • $\begingroup$ Your $F_n$ is wrong, already for $n=4$. The correct one is the one I wrote. So $n=5$ and 6 cannot be discarded that way. $\endgroup$ – jj_p Oct 20 '18 at 17:46
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    $\begingroup$ Show me a configuration with $4$ points such the lines connecting two of them create more than $18$ regions. You probbly used the formula for $\displaystyle\binom{n}{2}$ lines in general position (where you can indeed create $\displaystyle\frac12\,\binom{n}{2}\,\Biggr(\binom{n}{2}+1\Biggl)+1$ regions), but these lines are not in general position. The $n$ points are, and each of these points will be the intersection of $n-1$ lines. $\endgroup$ – Batominovski Oct 20 '18 at 17:49
  • $\begingroup$ I see, you are right. Your estimate is sharper. $\endgroup$ – jj_p Oct 20 '18 at 17:56

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