Abigail, Bart, Carol, Dan, Ehud, Fred, Golda, Hiro, and Juan are to be seated at a circular table. As usual, two seating arrangements are considered different exactly when someone has a different person to his or her right. If Fred can not be seated next to Ehud or Golda, how many different seating arrangments are possible? Thanks in advance Solution It doesn\'t matter where the first person sits; that person is a frame of reference. Let\'s have Abigail be the frame of reference. Bart then has 8 possible choices for where he can sit. Carol has 7. Dan has 6. Ehud has 5. Fred (we\'ll get back to him later). Golda has 4, Hiro 3, and Juan 2. Fred then is left with one spot left. 2/9 of this time (on average), he will be next to Ehud, Golda, or both of them. So, if we take 8! then multiply that by 7/9, we\'ll get how many arrangements there are at the table total. 8! * 7 / 9 = 31360..

1. Abigail, Bart, Carol, Dan, Ehud, Fred, Golda, Hiro, and Juan are to be seated at a circular table.
As usual, two seating arrangements are considered different exactly when someone has a
different person to his or her right. If Fred can not be seated next to Ehud or Golda, how many
different seating arrangments are possible?
Thanks in advance
Solution
It doesn't matter where the first person sits; that person is a frame of reference.
Let's have Abigail be the frame of reference. Bart then has 8 possible choices for where he can
sit. Carol has 7. Dan has 6. Ehud has 5. Fred (we'll get back to him later). Golda has 4, Hiro 3,
and Juan 2. Fred then is left with one spot left. 2/9 of this time (on average), he will be next to
Ehud, Golda, or both of them. So, if we take 8! then multiply that by 7/9, we'll get how many
arrangements there are at the table total. 8! * 7 / 9 = 31360.