# How many seating arrangements can George make if rotations and reflections of each seating arrangement are not considered different?

I was wondering if I could get some help with this problem. I am a middle school student, so simpler explanations would be very much appreciated.

George is planning a dinner party for three other couples, his wife, and himself. He plans to seat the four couples around a circular table for 8, and wants each husband to be seated opposite his wife. How many seating arrangements can he make, if rotations and reflections of each seating arrangement are not considered different? (Note: In this problem, if one seating is a reflection of another, then the two are considered the same!)

I know that there would be (8-1)! total arrangements for a round table where rotations are considered the same. Am I supposed to divide by 8 again to account for the reflections?

• Note that $(8-1)!$ does not account for the requirement that each couple must be seated exactly opposite each other. – Henning Makholm Mar 24 '18 at 18:57

Let's number the chairs around the table:

     8
7       1

6         2

5       3
4


Since rotations don't matter, we can decide once and for all that George himself will sit in chair 8, and his wife therefore in chair 4.

Since reflections don't matter, we can decide once and for all that the person sitting in chair 2 must be a man.

Convince yourself that every allowed seating is a possibly-rotated-or-reflected version of exactly one seating that follows these two additional rule.

Now, we can start by choosing who sits in chair 1. There are 6 possibilities for that, and our choice will also implicitly put that person's spouse in chair 5.

Now there are 4 possibilities for who to sit in chair 3.

Finally only one man and his wife are left. By our decision the man must sit in chair 2, so there's no choice here.

All in all we have $$6 \times 4 = 24$$ possible seatings.

• After you pointed out the flaw in my attempt, I took a slightly different approach that gave me $6 \cdot 4 \cdot 2$ ways of seating the women before reflections are taken into account, confirming your answer. – N. F. Taussig Mar 24 '18 at 19:10
• Thank you for your help! This means that if we put a husband in chair 1, his wife is automatically put in the opposite chair, so we don’t need to count for the wife, right? – starsinajar Mar 24 '18 at 21:31

Seat George. His wife must be seated opposite him. That leaves six open seats. Line up the other women in some order. The first of them can be seated in six ways. Her husband sits opposite to her, leaving four choices for the next woman. Her husband sits opposite to her. That leaves two choices for the third woman. Her husband sits opposite to her. Therefore, up to rotation, there are $6 \cdot 4 \cdot 2$ possible seating arrangements. However, we have not accounted for reflections, so we must divide by $2$, which gives $$\frac{1}{2} \cdot 6 \cdot 4 \cdot 2 = 24$$ seating arrangements, in agreement with Henning Makholm's answer.

My thanks to Henning Makholm for pointing out the flaw in my initial attempt to solve the problem.

• Thank you for your answer! Do we divide by 2 because the arrangements are counted twice when we reflect? – starsinajar Mar 24 '18 at 22:33
• We divide by $2$ since the clockwise arrangement $ABCDEFGH$ becomes the clockwise arrangement $HGFEDCBA$ upon reflection. Each arrangement is considered to be equivalent to its reversal under reflection. The number of such pairs is half the number of clockwise arrangements up to rotation. – N. F. Taussig Mar 25 '18 at 0:48