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?