Two straight lines intersect at a point P . There are n distinct points on the first line (excluding P ) and 2n distinct points on the second line (excluding P ). If the total number of triangles that can be formed by choosing 3 points from these (3n+1) points is 375 , then the value of n is:

Step-by-step solution

Total number of points is n + 2n + 1 = 3n + 1 . The total number of triangles that can be formed is the number of ways to choose 3 points out of (3n+1) minus the number of ways to choose 3 collinear points. The first line contains n + 1 points (including P ). The second line contains 2n + 1 points (including P ). Total triangles = ^ 3n+1 C₃ - ^ n+1 C₃ - ^ 2n+1 C₃ = 375 . Expanding the combinations: (3n+1)(3n)(3n-1) 6 - (n+1)(n)(n-1) 6 - (2n+1)(2n)(2n-1) 6 = 375 n 6 [ 3(9n^2 - 1) - (n^2 - 1) - 2(4n^2 - 1) ] = 375 n