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?
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 $P$.