Quandaries
and Queries 

Who is asking: Parent Level: Secondary Question: The solution is (405)(52)(25)/405 choose 3 I just dont see how this is possible! 

Hi, First one must note that a diagonal here does not include an edge of the polygon. Thus the number of diagonals is 30C2 30 = 405. The number of ways to pick three diagonals in thus 405C3. To find the number of ways to get three diagonals through a point we first choose a point in 30C1 = 30 ways and then pick the other three end points of these diagonals; this can be done in 27C3 ways (you can't pick points adjacent to the one that you picked already as that forms an edge of the polygon, not a diagonal). The probability that three diagonals selected at random go through a common point is thus p = [30(27C3)]/405C3 = 30*27*26*25/405*404*403 = 2*26*25/404*403 Denis 

