急! F5 排列與組合 4q2

(23a) For n points, taking any 2 points will get one diagonal except for those that will be the sides of the polygon.
So no. of diagonals = nC2 - n = n(n - 1)/2 - n = n^2/2 - 3n/2
(23b) n^2/2 - 3n/2 = 54
n^2 - 3n = 108
n^2 - 3n - 108 = 0
(n + 9)(n - 12) = 0
so n = - 9 (rej.) or 12.
(25a) No. of ways = 388!
(25b) Possible combination = 388P5