What is the fastest way to solve this combinatorics problem?

Question

I have eight sets, $A_1$, $A_2$, $A_3$, $A_4$, $B_1$, $B_2$, $B_3$, $B_4$ and I know that:

• $|A_i| = |B_j| = 4$ for all $i, j$
• $|B_i \cap B_j| = |A_i \cap A_j| = 2$ whenever $i \neq j$
• $|A_i \cap B_j| = 3$ for all $i, j$

and I want to determine if this situation can actually occur, and if yes the possible number of elements of the union of all eight sets. Is there a fast way to approach this? (it is a sub-problem of a general problem I am working on, where $2$ could be replaced by $4$ for certain $i$, $j$, and $3$ could be replaced by $1$. I plan to consider all the cases and list how many elements the union of all of them has).

Answer 1

No, this is not possible. For any $i$, since $A_i \cap B_1$ and $A_i \cap B_2$ must share at least 2 elements -- they both have size 3, drawn from $A_i$ of size 4 -- we have that each $A_i$ must also contain the two elements that $B_1, B_2$ share. But then each $A_i$ must consist of those two elements, plus two elements which are not shared by any of the other $A_i$. By a symmetric argument, those two elements must also be in each $B_j$.

For $A_1$, call the two other elements $a, a'$. We have that precisely one of $a, a'$ is in each of the $B_j$, which means there must either be at least two $B_j$ containing $a$, or at least two $B_j$ containing $a'$. But this contradicts the pairwise intersection among the $B_j$ having precisely 2 elements.

Answer 2

We may as well let $A_1=\{1,2,3,4\}$, in which case we may as well let $B_1=\{1,2,3,5\}$, $B_2=\{1,2,4,6\}$, $B_3=\{1,3,4,7\}$, and $B_4=\{2,3,4,8\}$. Starting next with $B_1=\{1,2,3,5\}$, we may as well let $A_2=\{1,2,5,x\}$, with $x\gt5$. But now we're stuck, because $A_2\cap B_3$ can have at most $2$ elements. So the three criteria for the eight sets cannot all be satisfied.