How to find 31 different subsets of $[11]$, $S_1,S_2,\cdots,S_{31}$, so that $\forall i\in[31], |S_i|=5$ and $\forall i,j\in[31]$, $|S_i\cap S_j|\ge 3$?
My first attempt was to fix three common elements and then choose the remaining elements from the remaining part of $[11]$, which gives the result ${11-3\choose 2}=28$. I also tried some other methods but none of them gives an answer larger than 28...
Thanks in advance.
For each $i$ in $[5]$, let $A_i = [5]-\{i\}$. So for example, the set $A_2 = \{1,3,4,5\}$. Clearly each pair $i$ and $j$ satisfies $|A_i \cap A_j| \geq 3$.
Now for each set $A_i$ (of which there are 5), there are 6 distinct sets of the form $A_i \cup \{k\}$, where $k$ is an integer in $\{6, 7, 8, 9, 10, 11\}$. For example, $A_3 \cup \{8\} = \{1,2,4,5,8\}$
So our collection of 31 sets will be the 30 sets of the form $A_i \cup \{k\}$, $i \in [5]$, $k \in [11]-[5]$, and the 1 set $[5]$.