In how many ways can you seat $12$ people at two round tables with $ 6$ places each? Think of possible ways of defining when two seatings are different, and find the answer for each.
Attempt:
Two considerations:
- Do we count equivalent rotations?
- Does the table matter?
Case $1$: Rotations & table matters
There are $12!$ ways of seating the group
Case $2$: Rotations matter, table doesn't
There are two ways to switch the tables. So there are $\frac{12!}{2}$ ways to seat the group.
Case $3$: Rotations don't matter, table does
Consider seating relative to special guest. There are $11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 $ ways to seat people around the first table. Seat the next special guest. There are $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ ways to seat the rest. The total is $11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 + 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.$
Case $4$: Rotations and tables don't matter
$\dfrac{(11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 + 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}{2}$
Best Answer
We can choose the group that will be at each table: $\frac{C_ {12}^{6}}{2} $ ways
After that, consider a circular permutation of both tables: $5!5!$
Then there will be $ \frac{C_{12}^{6}}{2}5!5! = 6,652,800 $ ways
In case the tables are different, simply multiply the answer by 2.