Seating $2n$ people around a table - Why $(2n - 1)!$ and not $(2n)!$?

Question

There've been numerous questions about this so please let me know if this is a duplicate.

Page 12 in http://www.am.qub.ac.uk/users/g.gribakin/sor/Chap1a.pdf says:

Let $A(i, r) =$ couple $i_r$ sit next to each other.
To compute the generic term $P[ \, A(i, 1) \cap A(i, 2) \cap \ldots \cap A(i, r) \, ]$, we proceed as follows.
There are $(2n - 1)!$ ways of seating $2n$ people at a roundtable. Why? Put the first person on some seat, then arrange the other $(2n - 1)$ around them.

My first thought was that seating $2n$ people around a roundtable means selecting $2n$ seats without replacement and with ordering. There'd be $(2n)$! ways of doing so.

How and why is this wrong? Shouldn't both ways work? It seems more guileful to "put the first person ... $(2n - 1)$ around them"?

Answer 1

The thing you're missing here is that in these sorts of problems, we generally consider two arrangements "the same" if one is a rotation of the other. So, for instance, the arrangements (read, say, clockwise around the table) ABCD and BCDA are generally considered the same... each person has the same neighbors to their left and right.

So, we think of this differently: once we've placed one person, the rest of the people flow from there in a permutation. This is where $(2n-1)!$ comes from: that first person's seat doesn't matter, but once they're in place any arrangement of the other $2n-1$ is distinct.

Another way to think of this is as follows: if we count up all the permutations of $\{1,2,\ldots,2n\}$, then you can note that there are $2n$ rotations that all lead to the same round-table configuration. Thus there are $$ \frac{(2n)!}{2n}=(2n-1)!$$ ways to arrange them around the table.

If we care which particular chair each person lands in, rather than just what their arrangement is, then you are absolutely right that it would be $(2n)!$ instead of $(2n-1)!$.

Answer 2

Ok. Lets take a look at the case of $n=4$ people.

There will be seating arrangements

$A,B,C,D$

$B,C,D,A$

$C,D,A,B$

$D,A,B,C$

However, if you seat them around the table, the "start" and "end" of the arrangement doesn't matter anymore, so you need to divide by n=4 after you get the number of arrangements for a row.

Answer 3

As the table is round, no seat is distinct.

Let's say there's a table with seats S1, S2, S3... S2N. One possible seating of people P1, P2, P3...P2N is to put them in that order in corresponding spots. However, if everyone shifts one seat to their right, the seating is still the same because the seats are the same, but with (2n)! it would be counted as a different case. Therefore we must divide the cases by the number of 'rotations', in this case 2n.

(2n!)/2n=(2n-1)!

Answer 4

Let's suppose that there are $n$ people, as the $2n$ case follows directly. It's helpful to think of the people not as an ordering around a table but as a linear ordering i.e. in a line. In this fashion we can build an $n$ to $1$ mapping $\pi:A \rightarrow B$ from the set $A$ of linear orderings of the people i.e. permutations of the people, to the set $B$ of arrangements of $n$ people around a circular table. How is this mapping defined? Well consider an $n$ permutation, say $a_1a_2\ldots a_n$, and place each element around a circular table starting with $a_1$ at the top and continuing to place elements counterclockwise. (Just for reference, note that $a_2$ is to the immediate left of $a_1$ in at our table and $a_n$ is to the immediate right).

Now consider what happens to the permutation $a_2a_3\ldots a_na_1$ under this same mapping. Now $a_2$ will be at the top, but it should be clear that $a_1$ is to the immediate right, and in fact $a_n$ is to the immediate right of that. So under the mapping the elements are in the same relative position as when $a_1a_2\ldots a_n$ was mapped. It is not hard to see that for a given permutation $a_1a_2\ldots a_n$, there are $n-1$ other permutations which can be obtained just by shifting a selection of elements to the back, and each of these will map to the same element under $\pi$ (e.g. $a_1a_2\ldots a_n$, $a_2a_3\ldots a_na_1$,$\ldots, a_na_1a_2\ldots a_{n-1}$)

Thus for every $n$ permutations in a line we get 1 arrangement around the table. Since there are $n!$ linear permutations, there must be $\frac{n!}{n} = (n-1)!$ elements in $B$.

Answer 5

I thought to dilate on this part of Nicholas Petersen's sterling answer:
Why are there $2n$ rotations that all lead to the same round-table configuration for $\color{green}{\{{1,2,\ldots,2n\}}}$?

Define $S(i) =$ seat $i$. Then: $\{\underbrace{\color{#FF4F00}{1}}_{S(1)}, \underbrace{\color{green}{2}}_{S(2)},...,\underbrace{\color{green}{2n - 1}}_{S(2n - 1)}, \underbrace{\color{green}{2n}}_{S(2n)}\}$
$\{\underbrace{\color{#FF4F00}{2}}_{S(1)}, \underbrace{\color{green}{3}}_{S(2)},..., \underbrace{\color{green}{2n}}_{S(2n - 1)}, \underbrace{\color{green}{1}}_{S(2n)}\}$
$\vdots$
$\{\underbrace{\color{#FF4F00}{2n - 1}}_{S(1)}, \underbrace{\color{green}{2n}}_{S(2)},...,\underbrace{\color{green}{1}}_{S(2n - 1)}, \underbrace{\color{green}{2}}_{S(2n)}\}$
$\{\underbrace{\color{#FF4F00}{2n}}_{S(1)}, \underbrace{\color{green}{1}}_{S(2)},...,\underbrace{\color{green}{2}}_{S(2n - 1)}, \underbrace{\color{green}{2n - 1}}_{S(2n)}\}$

Around a table, all $\color{#FF4F00}{2n}$ permutations above lead to the same (seating) configuration so the orange provides the answer.