Three vertices of a n-gon

Question

We choose three vertices of a convex n-gon, which form a triangle. If the number of ways we can choose the three vertices so that no sides of the triangle coincide with any sides of the n-gon is 7n, then find n.

Answer 1

The complementary problem consists in finding the number of ways in which we can choose the three vertices so that at least one side of the triangle thus defined is the side of the $n$-gon.

For this you first choose one side of the $n$-gon ($n$ ways), then the third point among the remaining $n-2$ vertices. However this way, the triangles made up of two consecutive sides of the $n$-gon are counted twice, so we must substract the number of pairs of consecutive sides, i.e. the number $n$ of three consecutive vertices. Thus there are $n(n-2)-n=n(n-3)$ ways for the complementary problem.

For the general problem in question, you have: $$\binom n3 -n(n-3)=\frac{n(n^2-9n+20)}6=\frac{n(n-4)(n-5)}6\enspace\text{ways}.$$

Now if this number is $7n$, it implies $(n-4)(n-5)=6\cdot7$, whence $n-5=6$, or $n=11$.