If $12$ distinct points are placed on a circle and all the chords connecting these points are drawn, at how many points do the chords intersect?

Question

If $12$ distinct points are placed on the circumference of a circle and all the chords connecting these points are drawn, at how many points do the chords intersect? Assume that no three chords intersect at the same point.

A) $12\choose2$

B) $12\choose4$

C) $2^{12}$

D) $\frac{12!}2$

I tried drawing a circle and tried to find a pattern but couldn't succeed. for 12 points I found the answer to be $(1+2+....+9)+(1+2+3+.....+8)+....+(1)$ and the result multiplied by $2$. But I'm getting $296$ which is in none of the options. Can anyone help?

Answer 1

If you select any $4$ distinct points on the circle, you'd have one distinct point of intersection. This'll give you a nice little formula of selecting $4$ points out of $n$.

$$N={n\choose4}={12\choose4}=495$$

Answer 2

Let's follow the suggestion of @saulspatz by considering the problem for fewer points. Consider the diagram below.

The points on each circle have been chosen in such a way that no three chords intersect at the same point. Under these conditions, we can see by inspection that

\begin{array}{c c} \text{number of points} & \text{number of intersections}\\ \hline 4 & 1\\ 5 & 5\\ 6 & 15 \end{array}

This should suggest a formula for the number of intersections when we have $n$ points and no three chords intersect in the same point.

A chord is determined by two points of the circle. Two intersecting chords are determined by four points of the circle since the only way the chords can intersect is if we connect both pairs of nonadjacent points.