I am trying to calculate the number of intersections one would have in a Venn diagram with 8 overlapping circles but do not know where I would start.
Any help with the number and how you got there would be appreciated.
Edit: It could be a Euler diagram too - basically I'm trying to present the argument to a journal editor that the upset plot that I used is far superior to a Venn/Euler diagram because of the sheer number of intersections one would have to look at.
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$\begingroup$Calculating the number of total intersections among n Venn diagrams involves summing the number of ways 2,3, ..., n circles intersect each other. Ask yourself:
- How many ways do two circles overlap?
- How many ways do three circles overlap?
- ...
- How many ways do $n$ circles overlap?
The answer to each of these questions is:
- n choose 2 - n choose 3 - ... - n choose nTherefore, we can express the number of intersections among n Venn diagrams as a formula with the following summation:$$I(n) = \sum_{i = 2}^{n} \frac{n!}{i!(n-i)!}$$$$I(8) = 247$$So a Venn diagram with 8 overlapping circles has 247 intersections.
$\endgroup$ 1 $\begingroup$Number of overlappings of r circles out of n circles is given by combinations choosing r from n ,that is r choose n = binomial (n,r) = C(n,r) When you consider n number of circles or sets on venn diagram, Intersections of 2 sets that is numberof overlappings of 2 circles = C(n,2) Intersections of 3 sets = C(n,3) ......................... .......................... Intersections of n sets = C(n,n) Total number of intersections = C(n,2)+C(n,3)+.......C(n,n) = {C(n,0)+C(n,1)+C(n,2)+........C(n,n)} - 1 -n = 2^n - n -1 Here you can obtain C(n,0)+C(n,1)+...C(n,n) by substituting x =1 to the binomial expansion of (1+x)^n
Now in your case since n =8 Total number of intersections is 247.
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