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

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?

• 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 $(7-1)!$? Commented Dec 16, 2014 at 8:28
• @DipanjanPal Got it, Thanks! Commented Dec 16, 2014 at 9:03
• You may also consider reflections different. Commented Dec 16, 2014 at 12:33
• 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. Commented Dec 17, 2014 at 5:55
• @1110101001 I suspect because the body highlights the "how is this different from.." aspect; perhaps the title should be updated to reflect that? Commented Dec 17, 2014 at 14:04

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.

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.

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.

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.

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.

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 $\frac{(7-1)!}{2}.$

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 $(n-1)!$

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.

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.

• Could you suggest what number is 7 − 2 equal to? I remember factorials yet (namely, that 6! = 720), but already forgot subtraction. Commented Dec 17, 2014 at 10:19
• 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. Commented Dec 17, 2014 at 21:41
• at least you should correct that "5!=120" Commented Jan 18, 2020 at 14:57

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: $6!$.

• @OllieFord: $7 - 1 = 6$. Commented Dec 17, 2014 at 14:43
• Jesus. Time for more coffee. As you were! Commented Dec 17, 2014 at 14:44
• @OllieFord: I usually blame that sort of thing on posting from my phone. Commented Dec 17, 2014 at 14:46
• Yes! That was it! Commented Dec 17, 2014 at 14:50

I think this illustrates why word problems are hard for students! You really have to ask them in a way that makes it clear how the words are to be interpreted mathematically.

If the question was just about (a row of) 7 chairs then $$7!$$ would the likely answer. The word circular suggests that you probably don't care about the difference between to arrangements where one can be got from the other by everybody moving a certain number of places left or right. If we are talking about making necklaces with 7 different coloured beads we would probably not care if one was obtained from another by rotating the beads; but we might also not care of the necklace was flipped over as well.

These kinds of problems can be solved using elementary group theory (counting orbits of group actions using a formula often attributed to Burnside). But here we can easily tackle it by elementary counting arguments as other answers indicate.