How many configurations can be made arranging $n$ objects in a circular alignment?

Question

How many configurations can be made arranging $n$ objects in a circular alignment?

Solution: $$\frac{n!}{n} = (n - 1)!$$ since for each permutation there are $n$ equivalents.

I don't understand the solution, can someone explain?

Answer 1

You can shift each alignment "by one" $n-1$ times without changing the order (note that the arrangement is circular), so you have to divide by $n$.

The initial number of alignments (when ignoring the circular structure) is just the number of ways to order $n$ objects ($n$ possibilities for the first object, $n-1$ for the second one, ..., so $n!$ in total).

Answer 2

Imagine taking $4$ people, Adam, Ben, Claire and Dave. Make them sit down on a straight bench. How many different ways they can do this?

Well, on the leftmost place you can choose from $4$ people. For the next seat you can pick from $3$ people and so on. This gives $4!$ different ways to sit down.

Observe that in this case $ABCD$ is definitely different from $DABC$ right?

Now imagine asking them to sit down around a circular table. In this case you could think in a similar fashion, BUT the above two ways wont be different. So in case of a circular table the following cases are the same (or more formally called equivalent)

$$ ABCD, \quad DABC,\quad CDAB, \quad BCDA $$

Observe that I always just moved the last one to the beginning. This could be done in $4$ times since there are $4$ people.

So for each arrangement on the straight bench we got $4$ equivalent arrangements around the circular table making the number of different arrangements around the table $4$ times less.

Now if you have $n$ people you could arrange them on a straight bench in $n!$ different ways but when you would like to arrange the around a circular table you will be able to shift $n$ times all of which would give rise to the same order (read, say clockwise) so I have to remove these cases by dividing with the number of people that is $n$. Giving rise to the formula in your question.

Hope this helped :)

Answer 3

It all depends on what you count as an equivalent configuration.

You start with $n!$ configurations, corresponding to the $n!$ permutations of $n$ distinguishable objects.

But if configurations that are related by a rotation are counted as equivalent, then each equivalence class contains $n$ different permutations. So the number of equivalent configurations is $\frac{n!}{n}=(n-1)!$.

And if configurations that are related by a rotation or a reflection are counted as equivalent, then each equivalence class contains $2n$ different permutations. So the number of equivalent configurations is then $\frac{n!}{2n}=\frac{(n-1)!}{2}$.