Consider six people standing in a line.

- $\displaystyle P_1, P_2 , P_3 , P_4 , P_5 , P_6$

They can be arranged in $\displaystyle 6!$ distinct ways.

Now suppose we arrange these people around a circle. The following permutation are not unique.

- $\displaystyle P_1, P_2 , P_3 , P_4 , P_5 , P_6$
- $\displaystyle P_2, P_3 , P_4 , P_5 , P_6 , P_1$

As one can become the other form rotating the table.

To solve this problem to keep one person fixed and arrange the other around the fixed person. so there will be $\displaystyle (6-1)!$ unique arrangements.

I hope this clarifies where the $\displaystyle (n-1)!$ comes from.

Side note:

In particular cases you will not consider clockwise and anticlockwise wise circular arrangements as being unique. For example, consider arranging beads in a necklace clockwise and anticlockwise arrangements are not consider unique and you can "flip" the necklace around to go form one to another. So in particular cases you may need to use $\displaystyle \frac{(n-1)!}{2}$

Bobak