# eight people around square table

I would like you to help me with one example. There are eight people(A,B,...,H). In how many ways can eight people, denoted A,B,.....,H be seated around the square table? The result is $$10,080$$ possible ways. I've also found three possible ways how to solve this example but I still can't get it.

1. The first method is 8!/4. Why do we have to divide by $$4$$ ?

2. The second method is 7! x 2. Why do we have to multiply by $$2$$ ?

3. The last method is following: In how many ways can i split these eight people to four groups of two? The last method is clear to me, but I do not know how to calculate it. Can you help me ?

1. A square table has four sides, and evidently we are considering rotations of a solution to be the "same" solution.

2. We pick one special person to sit down on the north side of the table. Now rotations are no longer relevant, so there are $$7!$$ ways to seat everyone else. We multiply by $$2$$ because our special person has two places to sit on the north side.

3. This is a different question than the previous ones, because the pairs of people now need to sit at the table, which they can do in different ways.

vadim123 has provided a nice answer to your first two questions. Therefore, I will focus on the third one.

Line up the eight people in some order, say alphabetically. The first person in line can be matched with a partner in seven ways. Remove those two people from the line. The first person remaining in the line can be matched with a partner in five ways. Remove those two people from the line. The first person remaining in the line can be matched with a partner in three ways. Remove those two people from the line. The two people remaining in the line must be partners. Hence, there are $$7 \cdot 5 \cdot 3 \cdot 1$$ ways to form four pairs of people.

Place the first person in line and her/his partner on one side of the table. The remaining three pairs of people can be placed on the remaining three sides of the table in $$3!$$ ways. On each of the four sides of the table, the people can arrange themselves in $$2!$$ ways, depending on who sits on the left.

Therefore, the number of distinguishable seating arrangements (up to rotation of the sides) is $$7 \cdot 5 \cdot 3 \cdot 1 \cdot 3! \cdot 2^4 = 10,080$$