Why do we subtract 1 when calculating permutations in a ring? 
$10$ persons are to be arranged in a ring shape. Number of ways to do that is $9!.$

I wonder why we subtarct $1$ in all such cases.
I can imagine that if A,B,C,D are sitting in a row then B,C,D,A would give me a different combination but had they been sitting on a circular table then both the above combinations would imply the same thing. Same for CDAB or DABC.
But how does it all lead to $(n-1)!$ formula? Individual cases, I am able to imagine but overall how are we generalizing it?

Imagine that the table is arranged so that one of the seats is due north of the centre of the table. Seat the $n$ people around the table. Let $p_1$ be the person sitting in that north seat, and let the other $n-1$ be $p_2,p_3,\ldots,p_n$ clockwise around the table. Now rotate the table and the seats one place counterclockwise: $p_2$ is now in the north seat, and the others going clockwise around the table are $p_3,p_4\ldots,p_n,p_1$. Repeat: $p_3$ is now sitting to the north, and clockwise around the table we have $p_4,p_5,\ldots,p_n,p_1,p_2$. If we keep doing this, we bring each person in succession to the northern spot, and after $n$ of these rotations everything and everyone is back in its original position. From the standpoint of absolute compass directions we had $n$ different arrangements, but from the standpoint of the order of people around the table these $n$ arrangements were all the same. Thus, each cyclic permutation of the diners corresponds to $n$ different absolute permutations, i.e., permutations in which we care about the absolute seating position of each diner and not just who sits next to whom. Since there are $n!$ absolute permutations, there must be $\frac{n!}n=(n-1)!$ cyclic permutations.
Alternatively, you can think of it this way: given a seating of the diners around the table, you can designate any seat as the head of the table and list the diners clockwise around the table starting at the head. That gives $n$ different lists, but they all correspond to the same cyclic permutation of the diners. Since $n!$ different lists are possible, there must again be $\frac{n!}n=(n-1)!$ different cyclic permutations, each corresponding to $n$ of the lists.
