# Number of arrangements of 7 persons in a round table with 7 seats

How many arrangements of $7$ persons at a round table with $7$ seats do you have?

I know there are two approaches for this problem:

1) If seats are numbered, I have $P_7=7!$ dispositions. I'm sure about it.

2) If seats are not numbered, I should consider the positions of people compared to each other (So I have e.g. $\{1,2,3,4,5,6,7\}=\{2,3,4,5,6,7,1\}=...=\{7,1,2,3,4,5,6\}=\{7,6,5,4,3,2,1\}=...\{1,7,6,5,4,3,2\}$ cause people are arranged always in the same way).

My book says solution for this point is $\frac{P_7}{7}$ but I'm not sure it's correct. I think I should remove more arrangements, but I don't know the number.

Am I wrong or am I right?

1. Suppose Alexandra is one of the people. We seat her at the table. It does not matter where. We then proceed to seat people clockwise around the table, starting at her immediate left. There are six choices for the person to her immediate left, five choices for the person to the immediate left of that person, and so forth. Hence, there are $6!$ distinguishable seating arrangements.
2. We can seat seven people in $7!$ orders. However, since only the relative order of the people matters, the seating arrangements are invariant under rotation. Rotations to the right by $0$, $1$, $2$, $3$, $4$, $5$, or $6$ places all preserve the same seating arrangements. Hence, the number of distinguishable seating arrangements is $$\frac{7!}{7} = 6!$$
• I understand your first approach and thank you for your corrections and answer. What I can't understand is why you divide the Permutation for $7$. – Dear An Jul 29 '17 at 10:28
• I also prefer the first approach. As for the second approach, notice that for each seating arrangement, there are seven rotations that preserve the same relative order, namely rotations to the right by $0$, $1$, $2$, $3$, $4$, $5$, or $6$ seats. Hence, we must divide by $7$ so that we do not count these arrangements seven times. – N. F. Taussig Jul 29 '17 at 10:34