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This question already has an answer here:

Suppose that n lines in general position are given in a plane. (General position means that no two lines are parallel, and no three lines have a common point.) Into how many regions do they divide the plane?

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marked as duplicate by Paul Sinclair, user147263, user26857, Claude Leibovici, Antonio Vargas Oct 21 '15 at 7:52

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My attempt (it might be wrong) the $n$-th line cuts all the $n-1$ other lines. The total number of intersections are $$N=\sum_{k=0}^n (k-1) = \frac{n(n-1)}{2}$$

You can consider a graph $G$, whose vertices are the intersections and each of this vertices is adjacent to 4 edges ($N$ vertices and $2N$ edges).

We miss the unbounded planes. One way to fix this have to add two vertices 'at infinity' for each line, join those vertices (a picture will convince you that you get the same number of faces). We get $2n$ extra vertices and $3n$ extra edges. $V=N+2n$, $E=2N+3n$.

By [Euler characteristic for planar graphs]:(https://en.wikipedia.org/wiki/Planar_graph#Euler.27s_formula)

$$V-E+F = 2$$ $$F=\frac{1}{2} \left(n^2+n-2\right)$$

Disclaimer: There is probably an easier way to do this.

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