#2**+1 **

Each pair of circles intersect twice

So.....we have a set of n circles and we want to choose any two of them = C (n, 2)

And since each pair intersects twice, the number of intersection points = 2 C(n , 2) =

2 n! / [ (n - 2)! * 2! ] = n! / ( n - 2)! = n ( n - 1)

So.....The max intersection points of n circles = n(n - 1)

So..... 4(4 - 1) = 4 (3) = 12

Just as EP found !!!!

CPhill Jan 17, 2019