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How many ways seven people can sit around a circular table?

For first, I thought it was $7!$ (the number of ways of sitting in seven chairs), but the answer is $(7-1)!$.

I don't understand how sitting around a circular table and sitting in seven chairs are different. Could somebody explain it please?

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    $\begingroup$ Since the table is circular, the arrangement $1234567$ and $2345671$ are considered to be same, contrary to what happened in the case of linear arrangments. Can you now find it why is it $(7-1)!$? $\endgroup$ – Dipanjan Pal Dec 16 '14 at 8:28
  • $\begingroup$ @DipanjanPal Got it, Thanks! $\endgroup$ – omidh Dec 16 '14 at 9:03
  • $\begingroup$ You may also consider reflections different. $\endgroup$ – Ethan Bolker Dec 16 '14 at 12:33
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    $\begingroup$ How did this get so many upvotes? This is a standard circular permutation problem that has been covered uncountably many times both on MSE and around the web. Searching google for "how many ways to sit around a circular table" brings up thousands of results. $\endgroup$ – 1110101001 Dec 17 '14 at 5:55
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    $\begingroup$ @1110101001 I suspect because the body highlights the "how is this different from.." aspect; perhaps the title should be updated to reflect that? $\endgroup$ – OJFord Dec 17 '14 at 14:04

10 Answers 10

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In a circular arrangement we first have to fix the position for the first person, which can be performed in only one way (since every position is considered same if no one is already sitting on any of the seats), also, because there are no mark on positions.

Now, we can also assume that remaining persons are to be seated in a line, because there is a fixed starting and ending point i.e. to the left or right of the first person.

Once we have fixed the position for the first person we can now arrange the remaining $(7-1)$ persons in $(7-1)!= 6!$ ways.

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It depends on what you mean by "how many ways".

It's not unreasonable to count two seatings around the table which only differ by a rotation as "the same".

On the other hand, if the chairs and the view from the chairs are different, it might make more sense to count those seatings as different.

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You can also think of it this way. In a straight line (i.e. seating seven people in seven chairs next to each other), there are clearly $7!$ ways. But when they are joined in a circle, a rotation still counts the same way of seating everyone. You'll notice that there are $7$ possible rotations in this case (since seven chairs). So we partition the result from the straight line into $7$ groups. This is $7!/7 = 6! = (7-1)!$. This idea is also called a circular permutation.

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First one person sits: there is just one possible way for her to sit, since seats are identical. Now the remaining seats differ since a new person may sit to the right or left (clockwise/anticlockwise) of the first person, therefore there are $6!$ ways for $6$ people to be situated around the table (with one place already taken by the first person). Therefore there are $1 \times 6!$ ways for people to sit around a circular table.

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Before approaching this question is important to see the difference between two very similar problems. Seating $7$ people at a table where each seat is numbered, and seating $7$ people at a table where the chairs are not numbered. When sitting at a table with chairs that are not numbered we will want to begin by placing the first person. At first you may think you have $7$ places to seat him! But because the table is not numbered any place you decide to place him is actually identical. This is because if you rotate any circle, you can get to the same table. This is why their is no significance to where you place the first person. After you place the first person then your placings start having significance.

And if the chairs are numbered it is $7!$. As it is an identical problem to lining up $7$ people.

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Fix position of first person and now there are $(7-1)!$ total ways. But if you don't consider anti-clockwise and clockwise different than $\frac{(7-1)!}{2}.$

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If what chair you sit on matters, your value of $7!$ would be correct; but if it doesn't, then it doesn't matter what chair you consider the "first" chair, or equivalently, who sits in it, all that matters is where the remaining 6 people sit: $6!$.

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  • $\begingroup$ @OllieFord: $7 - 1 = 6$. $\endgroup$ – orome Dec 17 '14 at 14:43
  • $\begingroup$ Jesus. Time for more coffee. As you were! $\endgroup$ – OJFord Dec 17 '14 at 14:44
  • $\begingroup$ @OllieFord: I usually blame that sort of thing on posting from my phone. $\endgroup$ – orome Dec 17 '14 at 14:46
  • $\begingroup$ Yes! That was it! $\endgroup$ – OJFord Dec 17 '14 at 14:50
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Other answers give correct explanations, but lack one nice intuition. If by "the same positioning" we mean "each person has the same people on his/her left and right side" (which seems quite natural), then this definition of "sameness" actually implies that "rotated" seatings are the same. This very same definition of "sameness" works differently when the people in question sit in a row: then one person has noone on the left, and another one - noone on the right.

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Are we dealing with chairs or a circular bench? If we are using chairs, then the number of combinations is $(7-1)!$ as previously noted. However if we are using a circular bench, then the location of the first two sitters is irrelevant. Accordingly the next five sitters can be placed in $(7-2)!=$$720$ ways.

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    $\begingroup$ Could you suggest what number is 7 − 2 equal to? I remember factorials yet (namely, that 6! = 720), but already forgot subtraction. $\endgroup$ – Incnis Mrsi Dec 17 '14 at 10:19
  • $\begingroup$ I want to thank you for this answer. It is wrong, but thought provoking. it is still (7-1)! but it gets there backwards. with chairs the more people that sit down, the fewer chairs remain, but with a bench, the more people that sit down the more places to sit. $\endgroup$ – hildred Dec 17 '14 at 21:41
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Number the positions around the circle from $1$ to $n$. Cut the circle at any position, say $1$, and lay out the circle as a straight line. The permutations of the $n$ points are $n!$ in number. But for a circle you could have cut the string at any point and still obtained a permutation of $n!$. Therefore we divide by $n$ to get $(n-1)! $

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protected by Bruno Joyal Dec 24 '14 at 8:27

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