I came across this question on a book on combinations and permutations. Although it had a vaguely defined one step answer : $\textbf{$^8P_2 = 56$}$.
As far as I could understand, the above step gives the number of arrangements possible for 8 objects taken two at a time. But I don't understand how it gives the points of intersection.
Also, is there any other method to solve this kind of problems ?
Regards
[Math] the maximum number of points of intersection of 8 circles
combinationspermutations
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Best Answer
Two circles can intersect in at most two points. So if you have $8$ circles for every pair there can be at most $2$ intersections giving us the answer $2\cdot{8\choose 2} = 56$.
However this does in and of itself not show that this number can also be reached. There might after all are points where $ 3$ circles meet lowering the total number of intersections.
Fortunately it is easy to construct $8$ circles that have no triple intersections. For example you can place $8$ circles all with radius $8$ on the coordinates $(0,0), (1,0),\ldots,(7,0)$. All these circles clearly intersect twice with each other. But if there was some point $x$ where three circles intersected then the circle with radius $8$ around $x$ would intersect the $x$-axis in $3$ points, which is clearly absurd.