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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?

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  • $\begingroup$ $n-1$ people? What is $n$? $\endgroup$ – Thomas Andrews Nov 6 '15 at 16:10
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    $\begingroup$ Hint: Be adventurous and choose $p$ people for the first table in one go. Then go about the arrangements in each. $\endgroup$ – Macavity Nov 6 '15 at 16:12
  • $\begingroup$ @ThomasAndrews wait I'll edit $\endgroup$ – Stabilo Nov 6 '15 at 16:14
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    $\begingroup$ Presumably, the question reuires circular tables so that the only thing that matters is the order around the circle, not the exact seating. $\endgroup$ – Thomas Andrews Nov 6 '15 at 16:17
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    $\begingroup$ The answers all agree and are correct as long as $p \neq q$. If the tables are the same size, you might want to divide by $2$ because you can swap the two tables. If the tables are labeled, you don't, but if they are interchangeable you do. $\endgroup$ – Ross Millikan Nov 6 '15 at 18:09
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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.

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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

$$ \frac{(p+q)!}{pq} $$

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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 ${p+q \choose p }(p-1)! (q-1)!$

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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:

$$\binom{p+q}{p}\cdot\binom{q}{q}\cdot\frac{p!}{p}\cdot\frac{q!}{q}=\frac{(p+q)!}{pq}$$

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