Mathematical proof regarding circular permutations The question here is in how many ways can $n$ people stand in order to form a ring.
Now I understand the concept behind it. I have understood it like this:
$n$ people can be arranged in $n!$ ways. Now in each ring, if the ring is broken from a unique point and is straightened to form a row, it would give rise to a unique permutation of $n$ people.
Since each ring consists of $n$ people, it would give rise to $n$ different permutations which in turn means that there are $n$ distinct permutations corresponding to any single ring.
$\therefore $ The number of ways in which a ring can be formed by $n$ people is $\dfrac{n!}{n}=(n-1)!$
Now what I feel is that this explanation is not sufficiently rigorous mathematically. Please help in providing a proper mathematical explanation for it.
I'm sorry if the question is very basic but please help.
Thanks
 A: I would say this (not very different from your justification): suppose you enumerate the $n$ people from a specific person, whatever its position on the ring. Two different arrangements then correspond exactly to two different permutations of the $n-1$ remaining people. Hence there are $(n-1)!$ such arrangements.
A: A way to be a bit more formal would be to say:

*

*Take a circle and slice it in $n$ equal part named $p_1, \dots ,p_n$.

*You have indeed $n!$ ways to arrange the $n$ persons  in the $n$ parts of the circle.

*Now, you don’t want to distinguish the same arrangements rotating the circle.

*So given an arrangement, its $n$ rotated figures won’t be distinguished.

*Therefore the division by $n$.

The bold sentence is the important one. If you do take care of where each person is sitting at the table (I don’t want to be close to the lamp...), then this sentence has to be cancelled and the result is $n!$.
A: Let $x$ be the number of ways of arranging $n$ people in a ring of size $n$.
Now for each of those $x$ rings, if we break them at one unique position (out of $n$ unique positions), it will correspond to one unique permutation.
so if we break all $x$ rings at all $n$ positions we must get all permutations of $n$ if they were arranged linearly. $$ x\cdot n = n!$$ Thus $$ x =  \dfrac{n!}{n}=(n-1)!$$
