TITLE: How many maximum triangles can be made?
QUESTION [7 upvotes]: There are $8$ points on a plane no three are colinear how many maximum triangles can be made s.t no two triangles have more than one point in common.
Now I can choose $3$ points from $8$ points in $^8C_3$ ways and two triangles can have two points in common if I choose $5$ points make $2$ triangles out of it and that can be made in $^8C_6 \times ^6C_3 \times \frac 12$ ways. So the answer should be $^8C_3-(^8C_5 \times ^5C_3\times \frac 12)$.
Am I double counting anything?
After seeing one comment and thinking a bit I feel that method of complementation will be harder here and I am thinking about how many ways to draw a triangle instead of maximum how many triangles,
So another approach: I can choose three points from $8$ points and draw the 1st triangle then the second triangle can be drawn taking one point from the first(because we are maximizing) and $2$ others from the remaining $5$ points. So we have used $5$ points and drew $2$ triangles. Then we can draw atmost one more triangle. So $3$ is the answer.

REPLY [6 votes]: bof gives a great justification of why it is eight or nine with an example of $8$.
Here is why it can't be nine. If there were nine triangles, they would use $27$ points, so one point would have to be used at least $4$ times. Each of these four triangles creates two edges containing this point, so we have at least $8$ edges containing this point. But there are only $7$ other points there must be a duplicate edge.