Why placing people in a circle has (n-1)! distinct arrangements? Why placing people in a circle has (n-1)! distinct arrangements?
I saw in books something about a placing a 'pivot', 
But As i continued reading i saw something that really confused me. Once we can distinguish any particular position in the circle, the rotational symmetry is lost and there are n! distinct arrangements.
Say if one of the chairs is broken (and we choose to take note of this fact) then there are n! distinct ways n people can sit – pick one person to sit in the broken chair, then a person to sit to their right, and so on. 
What's actually the difference between the 'broken chair' and placing a 'pivot'? And why is 'the broken chair' has n! while the other method has (n-1)!?
 A: Think it in this way: the permutations of  $n$ person are $n!$, each permutation gives rise to $n$ identical patterns for the table composition ($n$ rotations of one chair at each step) so that you obtain the desired number: $n!/n=(n-1)!$.
A: When they refer to placing a pivot, they just mean picking a starting place from which to count seating arrangements.  You could think of it this way: We get to pick an arbitrary place for the pivot, but we don't get to pick an arbitrary place for the broken chair.
A: Here is a better answer.
First of all there are 2 ways to do it. If we consider every arrangement different then there is N! arrangements. 
However Most circle problems consider this. Let's say we have 4 people
Albert, Bob, Craig, Dan in a circle. Then there is N! permutations. However let's keep them all in the same order and move each over one seat. So they are all in the same order but moved over 1 seat. We can do it again, and again. In fact Albert can sit in 4 different seats with Bob, Craig and Dan next to him.
So if we consider the order of A-B-C-D the same no matter the starting seat then it is 4!/4. 4 is because the starting seat doesn't matter so we are eliminating 4 possible arrangements.
Or it is N!/N
