Let there be 9 fixed point on the circumference of a circle. Let there be 9 fixed points on the circumference of a circle. Each of these points is joined to every one of the remaining 8 points by a straight line and the points are positioned on the circumference so that at most 2 of the aformentioned lines meet in any interior point of the circle, as shown in the example below. How many intersection points are there?

Here is what I did:- To form a line we need 2 points but there are 9 so there are ${9\choose2}$=36 lines. Two lines form a point of intersection so to find no of points of intersection we apply ${36\choose2}$ i.e 630 points of intersections but the answer is 126.

The answer is $\binom{9}{4}=126$
Why? suppose you have an intersection point, by the hypothesis this point is the crossing of exactly two segments, these segments give us four of the fixed points.
conversely, given four points there is exactly one pair of lines which have those points as endpoints which intersect (think of the cuadrilateral formed by the four points, you want the diagonals).
Thus the sets of four fixed points and the intersection points are in one-one correspondance, but the number of subsets of $4$ fixed points is $\binom{9}{4}$
