The problem
There are 2 tables seating 6 people each. With 12 people, how many arrangements (with all 12 people seated) are necessary so that every pair shares a table for the same number of arrangements?
Example
Here's an example with 4 arrangements:
Arrangement 1:
1,2,3,4,5,6
and7,8,9,10,11,12
Arrangement 2:1,2,3,4,11,12
and7,8,9,10,5,6
Arrangement 3:1,2,9,10,5,6
and7,8,3,4,11,12
Arrangement 4:7,8,3,4,5,6
and1,2,9,10,11,12
1
and2
share a table 4 times.
1
and3
share a table 2 times.
1
and4
share a table 2 times.
...
11
and12
share a table 4 times.This is an invalid answer, because not all pairs share a table the same number of times.
My reasoning so far
In a single arrangement, a given person shares a table with 5 people out of 11 (other people). So, that person must sit with every other person 5 out of 11 times. So the answer must be a multiple of 11.
There are $\frac{\binom{12}{6}}{2}= 462$ unique arrangements, so that's an upper bound. (Division by two because the two tables are interchangeable).
(I'm also curious about the same problem, but with 2 tables of $n$ people)