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?
 A: 6 times 4 parabola-parabola intersections, 12 times 4 parabola-circle intersections, 3 times 2 circle-circle intersections.

A: All pairs can potentially intersect in $4$ points, except for the pairs of circles, which can only have two intersection points.

It follows that an upper bound on the number of intersection points is 
$$4{\binom{4}{2}}+4(4\cdot 3)+2\binom{3}{2}=78$$
Explanation:


*

*${\large{\binom{4}{2}}}$ pairs of parabolas, each pair with at most $4$ intersection points.

*$4{\,\cdot\,}3$ parabola-circle pairs, each pair with at most $4$ intersection points.

*$\binom{3}{2}$ pairs of circles, each pair with at most $2$ intersection points.
 

I see no reason why the upper bound is not achievable.
A: Total number of intersections of parabolas is 24. For any two parabolas there is 4 intersections. There is 4C2 ways to choose 2 parabolas out of 4. 4C2 x 4 = 24.
For a circle and a parabola, the maximum number of intersections is 4. 
Total number of intersections of circles and parabolas is 4x4x3=48
For any two circles, the maximum number of intersections is 2.
Total intersections of circles is 3C2 x 2 =6
Maximum total is 24+48+6=78
A: 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.
