For this, I actually wrote down each possible seat they could be seated. For example, A J __ __ ___ , __ A J __ __ ___ etc…
And it gave me something very big because each would be (3\times2\times1)\times2 . Why is this method inaccurate?
To permute objects linearly, you have n! ways to do so as you might be aware. In the case of circular permutations, however, we can imagine rotating the elements in the table such that:
a_1 \mapsto a_2, a_2 \mapsto a_3, and so on…
and still have the same configuration. Thus, to account for distinct permutations, we consider the number of clockwise/anti-clockwise rotations we can make such that we get the same configuration.
All this leads to that fact that circular permutation of n objects is, in fact, (n - 1)!.
In your solution, you’re not accounting for this.