Why is the number of ways to arrange n people in distinct ways around a circle $\frac{n!}{n}$

Question

First off, this sounds like permutations, and I'm unable to squeeze $\frac{n!}{n}$ out of the permutations formula.

Could someone break down how to get to that solution from either the Combinations or Permutations standard formulas?

Answer 1

It is indeed permutations. Designate one of the seats the head of the table. Then you can imagine stretching the table out into a line, with the head of the table the first seat in the line. Now there are clearly $n!$ ways to seat the people. However, when we wrap the table back around into a circle and remove any indication of which seat was the head seat, for $n=5$, say, we can’t tell the permutation $12345$ from the permutations $23451,34512,45123$, and $51234$: all $5$ of them result in each person having exactly the same left and right neighbors. The only difference is in which seat we designated the head of the table, $1,2,3,4$, or $5$.

In general, given any seating around the table, we can pick any one of the $n$ people to be the head of the table and ‘uncurl’ the table into a permutation of the $n$ people. Each seating arrangement therefore gives rise to $n$ different permutations. Since there are $n!$ permutations altogether, there must therefore be $\frac{n!}n=(n-1)!$ different arrangements.

Answer 2

Where you place the first person does not really matter as all options yield the same outcome (just rotate the circle). Once the first person is seated it comes down to the problem of placing n-1 persons on n-1 seats which has $(n-1)!=n!/n$ possibilities (can see it like the first person "broke" the circle and now you have a line with n-1 seats).

Answer 3

We note that $\displaystyle \frac{n!}{n}$ is in fact another way of writing $(n-1)!$. And in turn, $(n-1)!$ is used, instead of $n!$ as we do normally, because all cyclic permutations are equivalent around the circle.

Answer 4

First, how many ways to place $n$ points on a circle? This is $n!$.

Let $(a_{1}, a_{2}, \ldots, a_{n})$ be a valid permutation, then $(a_{n}, a_{1}, a_{2}, \ldots, a_{n-1})$, $(a_{n-1}, a_{n}, a_{1}, a_{2}, \ldots, a_{n-2}), \ldots, $ also work as well. For any valid permuations there are $n$ that are equivalent to each other, so, in order to remove the repetitions, we divide $n!$ by $n$.

Answer 5

Imagine $5$ persons who want to sit around a (round) table with 5 seats. Number 1 chooses his seat. He has 5 choices. Then number 2 chooses his seat. He has 4 choices. You go on until number 5 who has only 1 choice left. After this, everybody has chosen his seat. You have $5\times 4\times 3\times 2\times 1 = 5!$ possibilities. These are the permutations, and you have $5!$ of them.

But you actually don't care if number 1 is on one seat or the other. All you care about is the order of the persons. So if everybody moves to the right by one seat or two,... or four, it's actually the same situation. So you divide by $n$, which is the number of possible rotations of the table.

Answer 6

As a hint, take $n = 12$ and think about the difference between hanging a clock in the normal position (1 way) versus hanging it with any of the twelve numbers at the top (twelve ways).