How many ways can $p+q$ people sit around $2$ circular tables - of sizes $p,q$? 
How many ways can $p+q$ people sit around $2$ circular tables - first
  table of size $p$ and the second of size $q$?

My attempt was:


*

*First choose one guy for the first table - $p+q\choose1$.

*Then choose the rest $p-1$ people - $(p+q-1)! \over p!$.
So we have now $p+q\choose1$$(p+q-1)! \over p!$.
Is that correct so far? But what about the second table? How can I consider him?

Choose the $p$ people around the first table: $p+q \choose p$.
For each set of $p$ people, there are $(p-1)!$ permutations once the first is placed (I assume a circular permutation has no effect on order, but "reversing" has). Likewise, there are $(q-1)!$ permutations for the second table.
All in all, there are ${p+q\choose p}(p-1)!(q-1)!$ possibilities.
