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$18$ points are selected on the circumference of a circle, all of which are connected to each other by straight lines.

If no three lines intersect at a common point,

What is the number of a triangles whose three vertices lie inside the circle?

If this question was asked for $6$ points, the answer would be $1$ as shown below:

enter image description here

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  • $\begingroup$ Does a triangle still count as a triangle, if it's divided by some other line(s)? $\endgroup$ – Bass Aug 17 '18 at 10:38
  • $\begingroup$ @bass yes. usual triangle count. $\endgroup$ – Oray Aug 17 '18 at 10:53
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Choose any $6$ of the $18$ points. Take the lines that connect pairs of opposite points of those six - this is the only set of lines on those points that will mutually intersect. Those intersections form an interior triangle.

Conversely, take any interior triangle. Consider the end points of the lines that form its sides. Those must be $6$ distinct points, otherwise the triangle would not be interior. Also, each line intersects the other two so each line will have two points on either side. This means that each line connects two opposite points of the six.

So there is a one-to-one relationship between the interior triangles and sets of $6$ points. We merely have to count how many ways there are to choose $6$ of the $18$ points to get the number of triangles.
There are ${{18}\choose{6}}=18564$ distinct interior triangles.

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  • $\begingroup$ The nice thing about this is that it is well illustrated in Oray's diagram so I found it easy to follow. $\endgroup$ – hexomino Aug 17 '18 at 8:54

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