I have looked around at MSE and I can't find any explanation to what I'm looking for.
I'm trying to understand the second reasoning to the formula $P_c = (n-1)! = \frac{n!}{n}$, where $P_c$ is the number of circular arrangements. Let me clarify what I mean by the second reasoning. The first one I've read is that a when a person sits down in a circle, there are $n$ indistinguishable positions, so in that sense, he only has $1$ relative position that he can sit in, the next people would have $(n-1)(n-2)\cdots1$.
Let me quote the second reasoning:
"Hence if we have ‘4’ things, then for each circular-arrangement number of linear-arrangements =4.
Similarly, if we have ‘n’ things, then for each circular – agreement, number of linear – arrangement = n."$^{[1]}$
Is there any better explanation to this; I'd like to understand the reasoning, but I'm just stuck in my own thinking, and it's blocking me from seeing what's, probably, obvious to most.
To add on this; how would you go about thinking if two people $A$ and $B$ were forced to sit next to each other, if $B$ sat to the right of $A$ or to the left are two permutations of course. And say for example there are $4$ other people there. How many circular permutations would that be?
The second reasoning is as follows. Every permutation of $n$ elements gives us some circular arrangement. Each arrangement is described this way $n$ times, since all rotations of a permutation correspond to the same circular arrangement (for example, 1234, 2341, 3412, 4123 are all the same circular arrangement). There are $n!$ permutations and so $n!/n = (n-1)!$ circular arrangements.