If there are n points and we pick just one, then there are n-1 connections or lines to the other points. Now take one of these so, excluding the connection to the first point, there are n-2 lines; take another point and there are n-3 lines, and so on. The total number of lines is n-1+n-2+...+2+1. Take this sum in pairs: n-1+1, n-2+2, n-3+3, etc. Each sum comes to n. We have (n-1)/2 pairs of sums n so in total we have **n(n-1)/2**. This formula clearly works if n is odd, because n-1 is even; and it also works when n is even, because n-1 will be odd so we have (n-2)/2 pairs, giving the sum n(n-2)/2, plus the middle number=n/2: n(n-2)/2+n/2=(n/2)(n-2+1)=**n(n-1)/2**. If n=4 the sum is 4*3/2=6; if n=3 the sum is 3*2/2=3.