Indistinguishable Tables

How many ways can $8$ people be seated on two indistinguishable tables that sit $4$ people each provided that persons $P$ and $Q$ are not seated at the same table?

My current thinking is that I first choose $3$ people to sit at one of the tables and then force $P$ to sit at that table. Then the remaining $4$ people (including $Q$) sit at the other table. Since the tables are indistinguishable, I need to divide my answer by $2$

$$\frac{{6\choose3}\times 3!\times 3!}{2!}=360$$

But I think this answer may be wrong, since the total number of ways of sitting $4$ people at each table with no restrictions is

$$\frac{{8\choose4}\times 3!\times 3!}{2!}=1260$$

But I intuitively feel that this should be double the previous result...

Can anyone see where my logic is not correct?

• Do we care about where people sit at a table? Is a table with $PABC$ clockwise different from a table with $PACB$? How about symmetry, is a table with $PABC$ clockwise different from one with $PCBA$ clockwise? Once we have a clear question there will be an answer. Your second is clearly wrong as it counts cases where $P$ and $Q$ are at the same table. – Ross Millikan Feb 27 '18 at 4:50
• Just call one table $P$ and the other one $Q$, then sit the remaining $6$ people at the two tables. – dxiv Feb 27 '18 at 4:50
• @RossMillikan Yes I care about those. This is why those $3!$ are in the answer. – Hugh Entwistle Feb 27 '18 at 4:52
• @dxiv If I do that, I still get the same answer – Hugh Entwistle Feb 27 '18 at 4:54
• 1260 minus the number of ways that they sit together? This gives $1260-540$ – WaveX Feb 27 '18 at 4:57

Your first answer is almost correct. Seating of $P$ and $Q$ leaves three distinguishable seats at each table. Once you seat them, you have $6 \choose 3$ ways to choose the people at $P$'s table and $3!$ ways to order them, then $3!$ ways to order the ones at $Q$'s table. There is no reason to divide by $2!$ because you can't swap the tables after you seat $P$.