I thought there would be 6 different ways where the 5 could sit without Bill and Dylan being next to each other, and then I multiplied 6 by 2 because Bill and Dylan could swap seats.Then I multiplied it by 6 again because there are 6 different ways in which Alexia, Caitlin and Evelyn could sit.
Well, we know that the number of permutations around a n-seater circular table is $(n-1)!$. For, $n=5$, we have a total of $24$ ways.
Now, to tackle the question, we go the opposite way, that is, we find the number of arrangements where Bob and Dylan are seated together, then subtract from above.
We tie Bob and Dylan as one unit, making a total of $4$ units, to be arranged in the circulation table. From above, we have a total of $(4-1)!=3!=6$ ways. But Bob and Dylan can themselves br arranged in two ways in the tie-up (B-D and D-B). So, we have a grand total of $6\times 2 =12$ ways.
Thus, the required number of ways equal $\boxed{24-12=12}$ ways. Hope it helps.
Notice that $B$ and $D$ cannot be seated next to each other $\Rightarrow$ it divides the group into 3 packets $(...)B(...)D(...)$ or like this $(...)D(...)B(...)$
But since it is a circular table we can consider that all people not between $B$ and $D$ are on the left (which is the same as the right) like this $(...)B(...)D$
Now let's consider the number of people seated between them, we have $C_3^k$ choices, but on the left there is no choice (people are determined automatically these are the remaining).
When we sum all this $=(2)(3\times1)+(1)(3\times2)=6+6=12$.
The order of $B,D$ doesn't matter because it is circular. If they swap, it will simply be represented by another (already counted) configuration with $B$ is before $D$.
Example : $(A)B(CE)D \leftrightarrow (A)D(CE)B=(CE)B(A)D$
Your first 6 should have been a 5 (maybe you forgot that positions 1 and 5 are next to each other because the table is circular?), so you get 60 instead of 72. Now the question is not really clear on what is supposed to be counted as different. If you consider two seating arrangements the same if they differ only by a rotation, then you have to divide by 5 and arrive at 12. So this is probably what was intended.
Yes, your idea is correct, but remember, the people are sitting around a circular table, which can be rotated. For example, if we were to arrange 6 people around a circular table, We wouldn't get $5! = 120$ ways, but instead, as you can place one person somewhere, and just focus on the people relative to his position. This means that, because there are $5$ different positions, we must divide by $5$, so $\frac{120}{5}= 24$ ways to arrange the positions.
Because it is a circular table, in our circle, all we need to focus on is one person. Let's say he's Bill.
Dylan, as he cannot sit next to Bill or on Bill's seat, has $5-2-1 = 2$ different ways to sit. Because of the circular table, Bill and Dylan don't need to switch seats.
And then, as you said correctly, Alice, Caitlin, and Evelyn can be ordered in $6$ different positions, and $2\cdot6 = 12.$