How many ways are there to cut a convex ( $n+2$ )-gon into triangles with diagonals?
#
Let's establish a one-to-one correspondence between the triangulations of a polygon and the placements of parentheses in the product $x_{0} x_{1} \ldots x_{n}$. We will highlight one of the sides $A B$ of the $(n+2)$-gon; all other sides will be numbered in order by the symbols $x_{0}, x_{1}, \ldots, x_{n}$. Each diagonal drawn "cuts off" a sequence of sides from the polygon that does not include $A B$. The corresponding part of the product $x_{0} x_{1} \ldots x_{n}$ will be enclosed in parentheses. In total, $n-1$ pairs of parentheses will be placed.

For example, the dissection of the hexagon in the figure corresponds to the placement of parentheses $\left(x_{0}\left(\left(x_{1} x_{2}\right) x_{3}\right)\right) x_{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_99ef25effe2e32f29eccg-05.jpg?height=452&width=443&top_left_y=1375&top_left_x=819)

It is easy to see that this method yields all possible placements of parentheses in the product.

## Answer

$C_{n}$.