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Find the number of ways $2n$ persons can be seated at $2$ round tables, with $n$ persons at each table.

Can anyone tell me how to solve this types of problems? Thanks for your help.

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The usual convention with a single round table is that two seatings of $n$ people are to be considered "the same" if they differ by a rotation. Thus the number of ways to seat $n$ people is $n!/n$, that is, $(n-1)!$.

The problem about $2$ round tables is a little ambiguous. Are the tables to be considered as (i) distinguishable (a red table and a blue one) or (ii) indistinguishable?

We first solve Problem (i). The people to be seated at the red table can be chosen in $\binom{2n}{n}$ ways. For each of these choices, the people can be arranged around the two tables in $((n-1)!)^2$ ways. Multiply. We can simplify quite a bit, for example to $\frac{2(2n-1)!}{n}$.

For interpretation (ii), divide the answer of (i) by $2$.

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How many ways can you split $2n$ people into two groups of size $n$?

Once they are split, how many ways are there to arrange $n$ people at a round table?

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