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- 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?

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    Note: A ... G are referring to the 7 people that want to sit around the table. – 2017-02-15
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    Are the chairs distinguishable? (e.g., are they numbered from 1 to 7?). – 2017-02-15
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    No they are not in my own example. The thing is, I want to know how I should answer table questions in different situations like chairs with numbers, without numbers, left to right and etc. – 2017-02-15
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    You need to decide if the specific chairs are indistinguishable (not labeled), or if the seven chairs are labeled: 1, 2, 3, 4, 5, 6, 7. – 2017-02-15
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    @amWhy, what if the chairs are not labeled, what happens with the arrangement ? – 2017-02-15
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    @Ciruss Even if two chairs are not labeled .. they are still different chairs. But maybe think about it the way I put it in my answer: if everyone shifts 1 seat clockwise, is that considered a different arrangement (since they are all sitting in a different chair now?) ... or is that the same arrangement? And if you say: how should I know? I say: exactly! That's a good thing to clarify! – 2017-02-15
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    Yes it is the same arrangement to me, because everyone is going from a fixed starting point (a random chair) to the end point (last chair) and knowing the number does not matter, so 6! – 2017-02-15
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    @Ciruss I would agree with that ... especially because as such the round table scenario would be different from 'straight line of chairs' scenario. But do we *know* that this is what the question meant? So, to me, the moral is this: it is *always* good to ask these kinds of clarification question! And don't think that you are *supposed* to know these kinds of things! So just ask the professor or whoever put this question to you. And if you don't get a straight answer, I would say to make the different interpretations explicit, and answer according to each one. – 2017-02-15
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    Yes you are right @Bram28, the best way to approach these questions is to ask the prof, if no answer then answering depending on ones own interpretations. – 2017-02-15

2 Answers 2

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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 ...)

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    Best to answer when OP has clarified (if not clear,then ask: best done in the comments below the question.) Secondly, before you answered, the OP replied to my *questions in comment section below question* and the OP replied, seemingly wanting to know how they are to deal with 7 different people (order matters), chairs indistinguishable, and with 7 different people (order matters) in 7 distinguishable (labeled) chairs... Just please refrain from "if you mean A, then B." "If you mean C, then D", and "If you meant EFGHIJK, then everything I just said doesn't apply" - sorts of questions. – 2017-02-15
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    Thank you for your motivation and explanation. I did not understand this part: "if it does not matter, you get half as many, i.e. (n−1)!/2(n−1)!/2". Why half ? Shouldn't you still get the same number ? – 2017-02-15
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    @Ciruss OK, consider an even simpler example where you have just 3 people. Now, if we go clockwise, and start with A, we can have either ABC or ACB, so 2 possibilities. But if clockwise and counter-clockwise arrangements (i.e. each other's mirror arrangement) are considered the same, then ABC and ACB are the same arrangement, so you have just 1 arrangement. So if clockwise vs counterclockwise does matter, you get (3-1)! = 2! =2 arrangement, but if mirror arrangements are considered the same, then you have just (3-1)!/2 = 2!/2 = 2/2 =1 arrangement – 2017-02-15
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    @Ciruss Here is another way to think about it, going back to the original 7. The way you calculate it, you say: OK, place person A in any of the seats. Then, in the seat clockwise from A, we can put one of six people, in the next one of 5, etc. So that gives 6! But then you say: if I go counterclockwise, and I do that again: place 1 out of 6 counterclockwise from A, 1 out of 5 counterclockwise from that, etc. So that's *another* 6!, hence 2*6! The problem is: when you were done doing all possible seatings clockwise, you *thereby* have covered all possible seatings counter clockwise! .... – 2017-02-15
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    @Ciruss ... That is, when counterclockwise I seat the people ABCDEFG, then that is the same arranegement as counterclockwise seating AGFEDCB. So, you should not multiply by 2 at all. Moreover, if mirror arrangements are considered the same, then a clockwise seating of ABCDEFG ends up being the same arranegemnt as the clockwise seating of AGFDECB. So: if mirror arrangements are consiered the same, you should divide the number by two. – 2017-02-15
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    One moment please, I am reading and trying to understand the logic. It is starting to make sense... – 2017-02-15
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    Ok I understand everything about CW and CCW rotation and that their arrangements are the same, so it is NOT necessary to times the arrangements by 2. I just want to know how can I identify if the question is meant to have mirror arrangements ? Should it specify in the question ? – 2017-02-15
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    @Ciruss It would have been best if the question would have specified that, but obviously it didn't. But it also didn't specific whether everyone shifting 1 seat would count as a different arrangement or not. In the latter case, we used our intuition, and concluded: well, that would probably be considered the same arrangement. So, what does our intuition about CW vs CCW? Personally, my intuition says that they are different ... but obviously I can't mind-read the author's intent! – 2017-02-15
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    Ok so I would rely on the text. @Bram28, thank you so much for your help my friend. You cannot believe how thankful I am to see someone spending so much time for free to help out the community, I'll definitely make sure to give back. – 2017-02-15
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    @Ciruss You're welcome! Share knowledge and spread good will indeed! :) – 2017-02-15
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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.

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    Yes, that is correct ... but it doesn't mean that you get 2*6! in the original scenario. You get the 2 factor in this case since we have 2 choices of where to place B relative to A .. you have nothing like that in the original scenario. In fact, think about it this way: by restricting B to sit next to A, you restrict B to 2 possible seats, while C can go in one of the remaining 5, D in one of 4, etc. So: 2*5*4*3*2*1 But if B can sit anywhere relative to A, then after placing A, B can go in one of 6 seats, after which C goes in one of 5, etc. So: 6*5*4*3*2*1 = 6! ... which is *not* 2*6! – 2017-02-15
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    I integrated these logical problems from 2 different, but also similar questions out of my textbook. In the question it specifies that B is restricted to sit beside A. So your first approach is right 2*5*4*3*2*1 or 2(5!). However, you are still right if the question did not limit the placing of A and B. Then that case 6! is right. – 2017-02-15