# Thread: Combinatorics - n straight lines?

1. ## Combinatorics - n straight lines?

Out of n straight lines whose lengths are 1, 2, 3,...,n inches respectively. Prove that the number of ways in which four may be chosen which will form a quadrilateral in which a circle may be inscribed is $\frac{1}{48}\{2n(n - 2)(2n - 5) - 3 + 3(-1)^n\}$.

2. ## A restatement

A quadrilateral with side lengths $a,b,c,d$ can be inscribed by a circle if and only if $a+c=b+d$. So to restate the problem in much simpler form, given the set of numbers $1,2,3,4,5,...n$, in how many ways can four distinct numbers be chosen from this set satisfying $a+b=c+d$ without regard to order?

Denote the sequence in question by $a_n$ which we are conjecturing is $1,3,7,13,22,34,50,...$