Permutations of Intersections of Parabola and Circles What is the maximum number of points of intersection for four distinct parabolas and three distinct circles drawn on a sheet of paper?

Anyone good enough to solve this.?

Well considering the two different parabolas can intersect at 4 different points. WHen you bring a third one in, its even more.

Two circles intersecting bring 2 points. A circle and a parabola brings 4 I guess. So how many in total?
6 times 4 parabola-parabola intersections, 12 times 4 parabola-circle intersections, 3 times 2 circle-circle intersections.
Four parabolas can have a maximum of 24 points of intersection.  A circle can have four points of intersection with a parabola and hence 16 with four parabolas. Three circles can have have 48 points of intersection.

Each circle can intersect other circle at 2 points and three circles can intersect with each other at 6 points and hence I revise it to be 78.