Discrete Math: Seating at a circular table - Possible Problem and My thinking:
Imagine a circular table, and you want to sit 7 people around it. The total arrangements would be 7!/7 or 6!. So, the order of left or right does not matter because we can sit these people anywhere and in any direction we want.
However, If they are denoted A, B, C, D, E, F, G. Then the order matters. Thus it would be 2*(7!/7) or 2*(6!).
- Logical Questions:
Am I right in my logic? How should I know the order of the left and right matters?
 A: You are right to have questions about this question: what exactly would make one arrangement different from another?  Does going clockwise constitute a different arrangement from going counter-clockwise? In fact: if everyone shifts a seat to the left, is that a different arrangement? Intuitively everyone shifting is not a different arrangement ... but the question just isn't clear on all this ... so you are right to be confused!
Also: these are good questions to have! I mean: not good for finding 'the correct answer' of course, but good for you and your critical mindset for asking these questions: that's exactly what good mathematicians do: it's not so much about the answer as it is about the analysis!
Now, what about your answer though? I mean: if this is for HW, how should you answer? Well, one thing that I think any self-respecting educator will appreciate is if you identify these different ways in which we can treat arrangements as the 'same' or 'different', and as such have a different answer for each specific interpretation of this question.
That said, you do seem to have a problem with your particular answer(s). That is, you say that if clockwise vs counter-clockwise does not matter, you get 6! possible arrangements (or: in general: with $n$ people you would get $(n-1)!$ arrangements, and if clockwise vs counter-clockwise does matter, you get twice as many.  Well, that's not right, since you are counting every arrangement twice. To give a simple example: if you have 4 people, then going clockwise you could have ABCD, or you could have ADCB, as 2 possible arrangements.  But if clockwise vs counter-clockwise does not matter, then these arrangements are in fact the same!  In other words, if clockwise vs counter-clockwise does matter, you get $(n-1)!$ arrangements, and if it does not matter, you get half as many, i.e. $(n-1)!/2$.
(And of course, these numbers are all assuming that everybody shifting 1 seat does not constitute a different arrangement. If it does, you have to put the $n$ back in ...)
A: Also I realized something:
If two people, assuming A and B individuals insist on sitting next to each other in a circular table of 7 people. Then now we should keep A and B together while we still can switch the places of C D E F G people. In another words the arrangements would be 2*(5!) because A to the right of B (BA) and B to the right of A (AB). The arrangements can go clockwise and counter clockwise. 
