Say I have two sets, A and B, that intersect. I know that the number of elements in either A or B, but not both, is x, and the intersection is y. Assuming the two sets are the same, I know that A and B have x/2 + y elements.
Now, let's say that I have A, B and C, and that I know that x is the number of elements in A or B or C, but not in any intersection, y is the total number of elements in the intersection of any two sets, and z is in the intersection of all three sets. Therefore, A and B and C have x/3+2 * y/3 + z elements.
I am trying to find the general formula for any number of equal sets where I only know how many elements are in either set, in the intersection of any two sets, or the intersection of any three sets and so on.
I know that the number of possible intersections of k sets out of n sets is the number of combinations, so my formula should be something like:
$\sum x_k * \frac{k!(n-k)!}{n!} * i_k$ where $i_k$ denotes how many times the set contains the intersection of k sets, and $x_k$ is the number of elements in the intersection of exactly any k sets out of the n sets. For example, in the case of three sets, A will contain two intersections of two sets (A$\cap$B and A$\cap$B).
So, for three sets I have: $x_1 * \frac{1}{3} + x_2 * \frac{1}{3} * 2 + x_3$
What would be the general formula?
The general formula is $${1\over n}x_1+{2\over n}x_2+{3\over n}x_3+\cdots+x_n$$ The coefficient of $x_k$ is $${_{n-1}C_{k-1}\over_nC_k}$$ where I'm writing $_nC_k$ for the binomial coefficient $n\choose k$.
The $_nC_k$ in the denominator comes from the number of intersections of $k$ of the $n$ sets, and the numerator comes from the same thing, only fixing one of the $k$ sets. Then a simple calculation shows the quotient of binomials reduces to $k/n$.