Permutations in a Circle calculations are a bit counter-intuitive. Imagine we have three people: Abe, Bob, and Cob. If they were sitting in a row of three seats, six distinct arrangements are possible:

$$ABC_1...ACB_2...BAC_3...BCA_4...CAB_5...CBA_6$$

And we can of course come to this number quite easily by simply doing the following:

$$3 \times 2 \times 1 =6$$

But What If...

...these three friends were sitting at a circular table with three seats around it? How many distinct arrangements would be possible? Can we simply do $3 \times 2 \times 1 =6$? The answer is no because, at a circular table, you get repeats. See below:

At a circular table these three cases are equal. I know what you're thinking. What do you mean equal? $A$ is at the top in the first case, $B$ is at the top in the second case, and $C$ is at the top in the third case.

But that's not what we mean. We mean distinct arrangements in terms of who is sitting next to whom. Notice that, in all three cases, $A$ is to the right of $B$ and to the left of $C$. Notice that, in all three cases, $B$ is to the left of $A$ and to the right of $C$.

So How Do We Get Rid of the Repeats?

We use this formula below. Assuming that there are $n$ people or items around a circle arrangement, we do...

$$(n-1)!$$