# How many ways can you seat $12$ people at two round tables?

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 $$\dfrac{(11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 + 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}{2}$$

• Have you learned Burnside's lemma? Oct 21, 2019 at 23:58
• Why addition and not multiplication in Case 4? And why division by $2$ in case 4? (Case 3 faces similar issues) Oct 22, 2019 at 0:13
• Traditionally, rotations of tables do not matter. Now, something that is rather ambiguous about this, do we consider the tables themselves different? Like, is there a red table and a blue table? Oct 22, 2019 at 0:17
• @JMoravitz Yes I'm considering different tables (colors/numbered) Oct 22, 2019 at 2:38
• So then, one of the people will be youngest. Let him sit at whichever table at whichever seat he likes, it matters not where so don't bother counting yet. Now, choose who sits to his right, then who sits to that person's right, etc... until the table is full. Now, among those people left, one will be youngest, let them sit at the remaining table wherever it matters not where. Then choose who sits to the right of that person and so on until all seats are full. That gives $11\cdot 10\cdot 9\cdot 8\cdot 7\times 5\cdot 4\cdot 3\cdot 2\cdot 1$ ways, not the middle is times, not plus. Oct 22, 2019 at 3:04

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