Let $\mathfrak{n} \geqslant 4$ be an integer. We draw all the diagonals of a convex n-gon. We assume that there are never three diagonals that intersect at the same point. What is the number of parts into which the n-gon is divided?

We will try to find the simplest formula possible.

## 2 Solutions
Count the regions that appear as diagonals are added. Initially, the n-gon without diagonals constitutes a single region. Then, when a diagonal is drawn, it is divided into 2. Therefore, we "create" one new region. We need to add $\frac{\mathfrak{n}(\mathfrak{n}-3)}{2}=\binom{\mathfrak{n}}{2}-n$ (the total number of diagonals) regions. This is true when the new diagonal drawn does not intersect any old diagonals. However, when two diagonals intersect, the number of additional regions "created" is 2, not 1. We have already counted one new region for these diagonals, so we need to count one additional region each time two diagonals intersect, which happens $\binom{\mathrm{n}}{4}$ times. Indeed, for each quadruplet $A, B, C, D$ of points on the polygon, there is a unique intersection of diagonals: if the points $A, B, C, D$ are located on the polygon in this order, it involves the diagonals $(A C)$ and (BD). Therefore, we have a total of

$$
1+\frac{\mathfrak{n}(n-1)}{2}-n+\frac{\mathfrak{n}(n-1)(n-2)(n-3)}{24}=\frac{n^{4}-6 n^{3}+23 n^{2}-42 n+24}{24}
$$

regions.

## 2 Group B

## 1 Statements