# How many ways are there to order in circle $n$ couples so each men sitsin front of his wife

I have the following question :

How many ways are there to order in circle $n$ couples so each men sits infront of his wife?

I thought of something like that :

Lets take wife $n$ and sit her down so the rest will be ordered according to her. now we have $n-1$ wifes and $n$ men, lets sit down all the wifes the left $n-1$ since we have a circle of $2n$ people ($n$ men, $n$ women) we need to sit down the rest of wifes in $2n-1$ sits (since the first women already sits) meaning choose $n-1$ wifes to sit in $2n-1$ sits, the rest of the husbands has one place to sit (infront of his wife)

I think my method is correct but when I choose $n-1$ from $2n-1$ could I get husband and wife instand of just the wifes? I mean I don't understand how choosing $n-1$ from $2n-1$ creates that I'll only choose wifes since if I also choose husbands I might take the husband with his wife and then I don't get that the rest has one place to sit.

Any ideas? Thanks!

• All answers seem to be assuming that you used "in front of" in the sense of "across from" and that arrangements related by a rotation are to be regarded as equivalent. Is this the intended interpretation of the question? – joriki May 23 '16 at 9:05

You're choosing $n-1$ seats from the $2n-1$ left, but this means you could be possibly choosing seats that are across from each other for women to sit in, which is not allowed under your restrictions.

Instead, let's pair off the husbands and wives and treat them as one item for now. We will first assume that the circle is entirely women on one half, and entirely men on the other half. Then fixing the first pair, there are $(n-1)!$ permutations for how the other women can sit around their half of the circle.

Now we have to consider that the man and woman can trade seats and still be across from each other. Since a couple either trades seats or doesn't trade seats, this gives $2^{n-1}$ options. Putting these two together gives a total of $2^{n-1}(n-1)!$ possible seating arrangements.

First Lady has $2n$ positions to choose. Now the position of her husband is fix (in front of her position). So next lady, say $2^{nd}$, has to choose a position out of $2n-2$. Once she choose her position, corresponding position of her husband got fixed. So, $3{rd}$ lady has $2n-4$ choices. Continuing this way we got total arrangements as :$$2n(2n-2)(2n-4)\ldots 2=2^n.n!$$ Since this is circular arrangement $2n$ of these arrangments are in fact same. So our answer will be: $$\frac{2^n.n!}{2n}=2^{n-1}(n-1)!$$

Fix one couple at the $12$ o'clock - $6$ o'clock diameter.

Each of the remaining couples can be fixed at $(n-1)$ diameters in $(n-1)!$ ways
with $2$ choices of pole for each diameter for (say) the wives,

thus $2^{n-1}(n-1)!$

Lets say there are $2n$ point on a circle So each women have $2n$ places to sit. The first women have $2n$ places to choose from, and for each place she chooses, her spouse will choose the place opposite of her.

So the next women have $2n-2$, and the next have $2n - 4$ and so on.

so the answer is $2n * (2n-2)*(2n-4) \cdots$

That is $2^n(n *(n-1)*(n-2) \cdots)$

$\Rightarrow$ $2^n*n!$

Kind of like this picture.