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?
Best Answer
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 $n$.
$$N={n\choose4}={12\choose4}=495$$