0
$\begingroup$

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.

$\endgroup$

2 Answers 2

2
$\begingroup$

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 $2$.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.