Consider 100 points in the plane, such that no three are collinear. We draw all the lines passing through two of these points, and we assume that if $A, B, C, D$ are four of these points, then $(A B)$ and $(C D)$ are not parallel, and their intersection point does not lie on another line. How many regions of the plane are thus formed by these lines?
We start with a region (the entire plane). Each time we draw a line, we add $1+m$ regions, where $m$ is the number of intersection points of the new line with the lines already drawn.

Let $n=100$. The number of regions is therefore equal to the sum of the following numbers:

$A=1$ (the number of initial regions);

$B=\binom{n}{2}$ (the number of lines drawn);

$C=n(n-2)$ (for each of the $n$ points, after drawing a line through this point, we draw another $n-2$ lines through this point);

$D=3\binom{n}{4}$ (if $A, B, C, D$ are four distinct points among the 100, then we form the new intersection points $(A B) \cap(C D),(A C) \cap(B D),(A D) \cap(B C))$.

The total number of regions is therefore

$$
1+\frac{n(n-1)}{2}+n(n-2)+\frac{n(n-1)(n-2)(n-3)}{8}=11778426
$$