Find the number of ways to seat seven people around a circular table if all rotations of a particular arrangement are considered to be the same.

Question

(a) Find the number of ways to seat seven people around a circular table if all rotations of a particular arrangement are considered to be the same.

(b) How many ways can seven keys be put on a circular key ring? (Note: The essential difference between keys on a ring and people around a table is that the keys will not object if the entire ring is turned upside down.)

(c) Suppose the key ring in (b) has a chain attached to it somewhere. How does that change the answer?

For a) my answer was $\frac{7!}{6}$ (this is what my professor got), but I don't think that's right. I think the correct answer is $\frac{7!}{7}=6!$.

For b) I am torn between $6!$ and $\frac{6!}{2}$.

For c) I have no idea how to answer this.

Can someone help me with these questions?

Answer 1

Considering the three different cases:

1. The number of ways to seat seven people at a round table equals: $$\frac{7!}{7} = 6! = 720$$ It does not matter where the first person is seated, as all rotations of a particular arrangement are considered to be the same. Therefore, we must divide by $7$.

2. The number of ways to put seven keys on a key ring equals: $$\frac{7!}{7 \cdot 2} = \frac{6!}{2} = 360$$ Not only rotations of a particular arrangement are the same, also arrangements which can be duplicated by turning the key ring are the same. For instance, the arrangement $A, B, C, D, E, F, G$ is the same as $G, F, E, D, C, B, A$ and $D, C, B, A, G, F, E$. Therefore, we must divide by $7$ and by $2$.

3. The number of ways to put seven keys on a key chain, with a chain on the ring, equals: $$\frac{8!}{8 \cdot 2} = \frac{7!}{2} = 2520$$ The key chain is an extra element which needs to be put on the key ring, resulting in $8!$ arrangements. Rotations of an arrangement are considered the same, and the key ring can again be turned upside down. For instance, the arrangement $|, A, B, C, D, E, F, G$ is the same as $G, F, E, D, C, B, A, |$ and $D, C, B, A, |, G, F, E$. Therefore, we must divide by $8$ and by $2$.

Answer 2

The answer for a is $6!$ First there are $7!$ combinations to sit the first person. This counts every combination $7$ times, as the number of rotations. You get $7!/7=6!$

You can view it from a different perspective. First, you sit the first person. There is one combination to do that since all sits are identical. Then, there are $6!$ combinations to sit the other people. Therefore you have $1 \cdot 6!$ combinations in total.