Number of intersections in a venn diagram with 8 circles 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.
 A: 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 n
Therefore, 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.
A: 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.
