Selection Examination A

(a) Any three from the $n$ points lying on the circle define a triangle. Therefore the number of triangles defined by the $n$ points is equal to the number of combinations of $n$ elements by $3$, that is $$\left(\genfrac{}{}{0ex}{}{n}{3}\right)=\frac{n(n-1)(n-2)}{6}$$ Therefore we get the equation: $\frac{n(n-1)(n-2)}{6}=2n\iff n=5$.

(β) Any four different points from the $n$ points lying on the circle define a convex quadrilateral whose diagonals intersect at an interior point of the circle and conversely. Therefore the number of interior points of the circle which are points of intersection of two chords is equal to $$\left(\genfrac{}{}{0ex}{}{n}{4}\right)=\frac{n!}{4!(n-4)!}=\frac{n(n-1)(n-2)(n-3)}{24}$$ Therefore we have the equation $$\frac{n(n-1)(n-2)(n-3)}{24}=5n\iff n=7$$