How many lines are determined by eight points if no three points are collinear? Also how many triangles are determined? I know this is the question of combinatorics.
So total number of points are eight so $n=8.$ If all points are non-collinear then total number of lines determined will be $^8C_8.$ Further I am thinking that in the given question three points are non-collinear so the total number of lines will be $^8C_3,$ may be. I don't get a point whether I am right or proceeding wrong. Secondly I don't know how to find the number of triangles.
Every pair of points determines a line, so that gives $\binom82=28$ lines. The question now is whether we have counted any line twice, and the the answer is "No," because there are no three of the points on any line.
Now for triangles, we have $\binom83=56$ ways to choose the vertices, and again the question is whether we have counted any triangle twice. Again, the answer is "No." If there were a fourth point in one of these triangles, it would lie on a side with two of the points, giving three collinear points.