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?

  • $\begingroup$ Try it for fewer points than $12$ and see if you can spot a pattern. $\endgroup$ – saulspatz Dec 12 '18 at 15:08
  • $\begingroup$ The numbers in (A)-(D) are all usually associated with counting certain (possibly ordered) subsets of a 12-element set. Can you tell what type of subsets these numbers count? Which type of subset corresponds to a single intersection point? $\endgroup$ – Mees de Vries Dec 12 '18 at 15:10
  • $\begingroup$ @saulspatz 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 $\endgroup$ – Ayaz S Imran Dec 12 '18 at 15:14
  • $\begingroup$ @MeesdeVries (A) is the number of intersection of 12 points $\endgroup$ – Ayaz S Imran Dec 12 '18 at 15:16
  • $\begingroup$ You should add your result, and the method you used to obtain it, to the body of the question. (Don't make another comment. Edit the question.) Then we'll be able to tell you where you've gone wrong. $\endgroup$ – saulspatz Dec 12 '18 at 15:16

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 $n$.



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.


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