> For example if we have 'n' objects then the first object could collide with n-1 objects (since we don't check if an object has collided with itself) the second object could collide with (n-2) additional objects not counting the possible collisions we have already counted. If we keep going like this we get the number of possible collisions is: (n-1) * (n-2) * (n-3) ... 1 This is equivalent to n! / 2! * (n-2)!

Seems to me like it should be more like the handshake problem, which comes out to n * (n-1)/2. n^2 is still bad, but not nearly as bad as n!.