8,9

How many parts do the diagonals of an $n$-gon divide it into, if no three diagonals intersect at the same point?
We will alternately draw diagonals. When we draw a new diagonal, the number of parts into which the previously drawn diagonals divide the polygon increases by $m+1$, where $m$ is the number of intersection points of the new diagonal with the previously drawn ones (the intersection points divide the diagonal into segments, each of which divides one of the old parts into two "new" ones). Thus, each new diagonal and each new intersection point of the diagonals increases the number of parts by 1. Therefore, the total number of parts into which the diagonals divide an $n$-gon is one more than the sum of the number of diagonals (which is $1 / 2 n(n-3)$, see problem $\underline{60391}$) and the number of intersection points of the diagonals (which is $C_{n}^{4}$, see problem $\underline{34981}$).

## Answer

Into $1 / 24 n(n-1)(n-2)(n-3)+1 / 2 n(n-3)+1$ parts.