Farrooh F. answered • 01/12/17
Tutor
5.0
(421)
Physics & Math Expert — AP, SAT, Olympiads, Advanced Courses
A pair of distinct lines (in plane geometry - Euclidean Geometry) can intersect at most one time. Using combinatorics you would have to solve for
101!/(99!*2!) = 101*100/1*2 = 101*50 =5050.
Alternatively, you can solve it this way. Line 1 crosses all 100. Line 2 does the same, but we already counted 1, so it does 99. And line 3 does 98, line 4 does 97, etc. Thus the sum of 1 to 100 is needed which is easily obtained by doing arithmetic progression: 5050.
101!/(99!*2!) = 101*100/1*2 = 101*50 =5050.
Alternatively, you can solve it this way. Line 1 crosses all 100. Line 2 does the same, but we already counted 1, so it does 99. And line 3 does 98, line 4 does 97, etc. Thus the sum of 1 to 100 is needed which is easily obtained by doing arithmetic progression: 5050.
Mark M.
01/12/17