In stead of lining $n$ objects, some problems involve circular arrangement. A typical case is to have $n$ people sit at a round table.  Two sitting plans are treated as the same if they are only different by rotation.

Such problem usually is not difficult to solve. The answer is $\boxed{(n-1)!}$ distinct sitting plans. There are two approaches to solve this problem:

• First discard the rotational congruent restriction. Then the first person has $n$ seats to choose from, the second person has $(n-1)$ seats to choose from, and so on. Hence the total number is $n!$. Now, let's consider rotation. For every such sitting plan, it can be rotated in $(n-1)$ different ways to become identical to another one. This means, the duplicate count is $n$ (including itself without rotation). Therefore, the final answer is $n!/n=(n-1)!$.
• Another way is to let the first person to sit on one seat. Because of the rotational congruent restriction, where he sits does not matter. Now use him as an anchor, arrange the rest $(n-1)$ people relative to his seat, there will be $(n-1)!$ different ways. Hence this is the final answer.

A variation to the above mentioned table arrangement is a bead necklace. A typical question will be how many different ways to use $n$ different colored beads to make a necklace? It is of course a circular arrangement. However, in addition to rotate the necklace, it is also possible to flip it. Hence, the answer will be $\boxed{\frac{(n-1)!}{2}}$.