$n$ points are distributed on the perimeter of a circle. We connect each $2$ points with a straight line, and we get the set $S$ of the chords.
(1) Find $|S|$
(2) Assuming that there are no inner point in the circle that lies on more than $2$ chords in $S$, how many intersection points there are between the chords in $S$ ?
My answer:
I don't know how to make it formal but I've found that the formula is $\frac{n(n-1)}{2}$ The only thing I can say that sounds formal is that we connect each point ($n$) with the other $n-1$ points so it's $n(n-1)$, but if we connect point $P_1$ with $P_2$ then we don't need to connect $P_2$ with $P_1$ again so we 'remove' duplicates by dividing by $2$
For (2) I have no idea where to start, I would appreciate your help on explaining how make (1) formal and how to get started with (2)
Thank you very much in advance