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?

There are $\binom{p+q}{p}$ ways to pick $p$ people to sit at the table that seats $p$ people (this automatically picks the $q$ people that sit at the other table since those are just the people that are left). For any one of these divisions, there are $p!/p = (p-1)!$ ways to order the people at the $p$ table (since it is circular) and similarly $(q-1)!$ ways to order the people at the $q$ table. To the solution is
$$ \binom{p+q}{p} (p-1)! (q-1)! $$
In fact, simplifying $\binom{p+q}{p} = \frac{(p+q)!}{p!q!}$ the solution can be written as
$$ \frac{(p+q)!}{pq} $$
