27.9*. In a convex $n$-gon $(n \geqslant 4)$, all diagonals are drawn. Into how many parts do they divide the $n$-gon, if no three of them intersect at the same point?
27.9. B. Let's draw the diagonals one by one. When we draw the next 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, i.e., 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 \(D+P+1\), where \(D\) is the number of diagonals, and \(P\) is the number of intersection points of the diagonals. It is clear that \(D=n(n-3) / 2\). According to the previous problem, \(P=n(n-1)(n-2)(n-3) / 24\).