SOLUTION: (a) In how many ways can the 10 kids in my class be seated in a circle if Aubrey and Fred insist on being seated diametrically opposite each other? (As usual, two seatings which

Question 1150826: (a) In how many ways can the 10 kids in my class be seated in a circle if Aubrey and Fred insist on
being seated diametrically opposite each other? (As usual, two seatings which are rotations of
each other are considered the same.)
(b) What if Betty and Guang also insist on being diametrically opposite each other?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Draw out the diagram of the 10 chairs seated around a circular table (optional)

Part (a)

Place Aubrey anywhere at the table. I'm going to have her seated at point A. Fred will then sit at point F which is diametrically opposite point A. In other words, segment FA goes through the center of the circle. Point F is the furthest point away from point A.

After these two seats are taken, we have 10-2 = 8 seats left with 8 people to fill them.

There are 8! = 8*7*6*5*4*3*2*1 = 40320 different permutations, so there are 40320 different ways to fill those 8 remaining seats.

If we placed Aubrey at point B, then Fred must sit at point G. There are 8! = 40320 ways to fill the remaining seats here; however, all we have done is just rotated each seating arrangment clockwise by one seat. This means that no new permutations have been introduced. Recall that the instructions state "two seatings which are rotations of each other are considered the same" which is another way of stating rotational symmetry. This logic applies if we placed Aubrey at any other seat (C, D, E, F, G, H, I, J) as well.

So effectively, we don't need to worry about where Aubrey is seated. All that matters are the 8 remaining chairs.

---------------------

Answer: 40320

====================================================

Part (b)

Refer to the diagram above if needed.

Place Aubrey at point A, so Fred goes to point F.

If we place Betty at point B, then Guang goes to point G.
However, we could place Betty at point C making Guang go to point H.
This is a different seating arrangement as there is one seat between Aubrey and Betty.

Betty has 8 choices on where to sit, once seated, there are 8-2 = 6 seats left to fill so there are 6! = 6*5*4*3*2*1 = 720 ways to fill them.

Overall, there are 8*6! = 8*720 = 5760 different seating arrangments if we have Aubrey opposite from Fred, and also Betty opposite from Guang.

---------------------

Answer: 5760
Note how this value is less than part (a)'s answer. This is because there are more restrictions on where Guang can sit. His seat is entirely dependent on Betty's seat choice (assuming she sits down before he does).

RELATED QUESTIONS
King Arthur's round table has 8 evenly spaced chairs. In how many ways can 8 knights be... (answered by CPhill)

In how many ways can 6 students be seated in a row of 6 seats if 2 of the students insist (answered by ikleyn)

In how many ways can 5 people be seated around a circular table if two of them insist on... (answered by math_helper)

how many was can 6 people be seated in a row if 3 of them insist on NOT sitting... (answered by Edwin McCravy)

How many ways can 6 people be seated in a row, if Ruth must be seated in the first chair? (answered by Edwin McCravy)

This is all about combination. 1. In how many ways can 6 students be seated in a row... (answered by psbhowmick)

Two freshmen, 2 sophomores, 2 juniors, and 3 seniors are on a panel. In how many ways can (answered by ikleyn)

How many ways can four couples be seated in a row of eight seats? (a)if couples are... (answered by ewatrrr)

a. Suppose we have a group of 8 students sitting on a long bench. Four of the eight... (answered by stanbon,edjones)