Both solutions are correct.
BUT they both give 100, not 120.
There are 10 points on a plane out of which 6 are collinear. How many triangles can be formed using these points as vertices.
I have tried it in two methods, and i getting different answers, i am not sure what is wrong in my approach.
10c3 - 6c3 = 100 i.e , taking three points from 10 and subtracting the chances in which all three points have been taken from the collinear points.
6c2 * 4c1 + 6c1*4c2 + 4c3 , i.e two from collinear and one from non-collinear + one from collinear and two from non-collinear + all three from non-collinear. I got 120.
I think my first approach has some flaws, can anyone identify them.