Problem on counting Let $S_1,...,S_{10}$ be 10 sets with exactly 8 elements, and $|S_i\cap S_j|\le 2$.
Problem: what is $\min(|S_1\cup...\cup S_{10}|)$
In general, if we have n sets $S_1,...,S_n$ with $|S_i|=N$ and $|S_{i_1}\cap...\cap S_{i_k}|\le m$ for every $\left\{i_1,...,i_k\right\}\subset\left\{1,...,n\right\}$, is there a way to find $\min(|\cup_{j=1}^n S_j|)$, or, to find an upper bound or lower bound for it?
Background: I am trying to prove that every group with 180 elements is not simple. By Sylow theorem, the number of Sylow-5 subgroups of a simple group with 180 elements can only be 6 or 36.If it contains exactly 6 Sylow-5 subgroups, then it can be embedding into $A_6$, which is impossible. Hence I tried to prove that it can not contain 36 Sylow-5 subgroups. I tried to make a contradiction by counting the number of elements. If it contains 36 Sylow-5 subgroups, it must contain 144 elements with order 5. And it turns out that such a group can only contain exactly 10 Sylow-3 subgroups. Let $S_1,..., S_n$ be these groups, if $|S_1\cup ...\cup S_{10}|\ge 37$, then the proof was finished. At first, I think the minimum takes place when $|S_1\cap...\cap S_{10}|=3$, but I can't prove it ( and it doesn't seem right ). To begin with the problem, I tried to consider the problem without group structure, and got the above problem.
 A: If I understand correctly, in order to get a contradiction, you would need to be able to prove that $|S_1\cup \dots \cup S_{10}|\ge 37$. However, the information $|S_i|=10$ and $|S_i\cap S_j|\le 2$ is insufficient to deduce this.
First, consider the below set system. Each $S_i$ in the below table is a subset of $\{1,\dots,13\}$ of size $4$, such that $|S_i\cap S_j|\le 1$. This can be modified to be $10$ sets of size $8$ with intersections of size at most $2$, by replacing each $i\in \{1,\dots,13\}$ with a doubled element, i.e,
$$
S_1=\{1,2,3,13\}\implies\{1_A,1_B,2_A,2_B,3_A,3_B,13_A,13_B\} 
$$
After making this replacement, we have $10$ subsets of a set of size $26$, each with size $8$ and at most $2$ elements in common, showing that you cannot conclude $|S_1\cup \dots \cup S_{10}|\ge 37$.
This example was derived from the finite projective plane over $\mathbb F_3$. I do not know whether something like this can actually arise in the group theory context you described.

A: Let $M = S_1\cup S_2\cup...\cup S_{10}$ and $|M|=n$. We see that each set $\{x,y,z\}$ is a subset of most one set $S_i$ and each $S_i$ contains ${8\choose 3}$ triple sets, so we have $$10\cdot {8\choose 3}\leq {n\choose 3} \implies (n-1)^3> 3360 $$
so $n\geq 16$.

But on the other hand we see that if we take $k$ sets from those 10 we have $$\Big| \bigcup _{i=1}^k S_i\Big| \geq \sum _{i=1}^k |S_i| - \sum _{1\leq i<j\leq k}|S_i\cap S_j|$$
so $$|M|\geq \max\{-k^2 +9k; \;\;k\in \{0,1,2,...,9\}\} = 20$$
so $n\geq 20$.
