PR 4 # 17
 

There are n points on a circle. A straight line segment is drawn between each pair of points. How many intersections are there within the circle if no 3 lines are concurrent?

[spoiler]n*(n-1)*(n^2-5n+6)/24

Pick a point - there are n ways to do that.

Consider sequentially the other points - there are n-1 considerations

For the "ith" other point, there are (i-1) intermediate points on one side and (n-i-1) points on the other side of the line. Thus there are (i-1)*(n-i-1) intersections of the line.

Every intersection has been counted 4 times - once for each endpoint of its lines.

Together this is

(n/4) sum From i=0 to (n-1) of (i-1)*(n-i-1)

Which simplifies to the above.

[/spoiler]

[spoiler](n^2-5n+6)[/spoiler] simplifies to [spoiler](n-2)*(n-3)[/spoiler]:razz:

That exposes a MUCH simpler derivation.

[spoiler]There are (n choose 4) ways to pick 4 points.
The lines through these 4 points have exactly one crossing,
and ever crossing corresponds to exactly one such set.
So the answer is (n choose 4)
[/spoiler]