A, B, C and D are seated around in a circular table. If B and C are not seated together, the number of arrangements are? 
A, B, C and D are seated around in a circular table. If B and C are not seated together, the number of arrangements are? 

I have read possible/similar duplicates to this question, however, they all have very long answers and I figured the working to this question should not be so complicated. 
Could someone please help figure out the most efficient way to work this out? 
 A: Unless otherwise specified, seats at a circular table are taken to be unnumbered, which means that rotation of any arrangement doesn't give a new arrangement.
Here the only way B and C can sit apart is diametrically opposite, so seat them thus.
A then has only 2 choices of seats, and D automatically sits in the last seat.
Thus answer is simply $2$ arrangements
A: HINT: First calculate the total number of possible seatings of the four people, without any restrictions. Remember that two seatings are considered the same if the four people are in the same cyclic order, so there are not $4!$ different seatings. 
Then you’ll calculate the number of unwanted seatings, i.e., seatings in which B and C are seated next to each other. Pretend that B and C are a single person, whom I’ll call E; how many ways are there to seat A, D, and E? Finally, remember that E isn’t really a single person: E is B and C seated side by side, and that can happen in two different ways, depending on which of B and C is on the other’s left. Thus, each of the ways to seat A, D, and E is really two ways to seat A, B, C, and D with B and C together.
Finally, subtract the number of unwanted seatings from the total number.
Added: As true blue anil points out, this problem is small enough to allow a very simple direct solution; I am giving you a general approach that works on all similar problems, irrespective of size.
