1. ## couples

How many ways are there to seat n married couples at a round table with 2n chairs in such a way that the couples never sit next to each other?

2. Edit: You may ignore this incorrect solution, I leave it for people to see where I did wrong.

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Firstly, we will find the number of the ways that the couples sit together.

As there are n couples, there will be (n-1)! different arrangements for the couples. And each couple can sit in 2! ways.

If a couple can sit in 2 ways, n couples can sit in 2^n ways.

Hence, n couples can be placed in 2^n (n-1)! ways. (when every couple sits together)

All arrangements - Couples sitting together = Couples not sitting together

We can place 2n people in (2n-1)! ways.

All arrangements - Couples sitting together

(2n-1)! - 2^n (n-1)!

3. Originally Posted by wingless
$\text{All arrangements}-\text{Couples sitting together}=\text{Couples not sitting together}$
It does not quite work that way.
The complement of “all couples together” is not “no couples together”.

What you want here is “no couples together” which is the complement of “at least one couple is together” .

4. Originally Posted by Plato
It does not quite work that way.
The complement of “all couples together” is not “no couples together”.

What you want here is “no couples together” which is the complement of “at least one couple is together” .
Oh, you're quite right! That was such a big, silly mistake.. So my solution is totally wrong