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$$(p+q-1)! \over p!$.
Is that correct so far? But what about the second table? How can I consider him?
 A: Choose the $p$ people around the first table: $p+q \choose p$.
For each set of $p$ people, there are $(p-1)!$ permutations once the first is placed (I assume a circular permutation has no effect on order, but "reversing" has). Likewise, there are $(q-1)!$ permutations for the second table.
All in all, there are ${p+q\choose p}(p-1)!(q-1)!$ possibilities.
A: There are $\binom{p+q}{p}$ ways to pick $p$ people to sit at the table that seats $p$ people (this automatically picks the $q$ people that sit at the other table since those are just the people that are left). For any one of these divisions, there are $p!/p = (p-1)!$ ways to order the people at the $p$ table (since it is circular) and similarly $(q-1)!$ ways to order the people at the $q$ table. To the solution is
$$ \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} $$
A: you can pick $p$ people to seat in the first table in $p+q \choose p$ ways. then there are $(p-1)! \times (q-1)!$ ways to seat on each table. therefore the total number of ways is ${p+q \choose p }(p-1)! (q-1)!$
A: Choose $p$ out of $p+q$ people to sit around the first table:
$$\binom{p+q}{p}$$
Choose the remaining $q$ people to sit around the second table:
$$\binom{q}{q}$$
Reorder the people around the first table, but ignore symmetric orders:
$$\frac{p!}{p}$$
Reorder the people around the second table, but ignore symmetric orders:
$$\frac{q!}{q}$$

The result is:
$$\binom{p+q}{p}\cdot\binom{q}{q}\cdot\frac{p!}{p}\cdot\frac{q!}{q}=\frac{(p+q)!}{pq}$$
