Selective Seating Arrangement
In how many ways can we seat 8 people around a table if Alice and Bob won't sit next to each other? (Two seatings are the same if one is a rotation of the other.)
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- $\frac{a}{b}$
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- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
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- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
There are 8 choices for seats for Alice to sit in. Once Alice is seated, there are 5 seats left for Bob, since he won't sit in either seat immediately next to Alice. This leaves 6 people to place in the remaining 6 seats, which can be done in $ 6! $ ways. However, we must divide by 8 to account for the 8 rotations of the table. So the number of arrangements is $ \dfrac{8 \times 5 \times 6!}{8} = 5 \times 6! = \boxed{3600} $. Alternatively, we can account for the rotations at the beginning, by fixing the table around Alice. Bob can't sit in her seat, or the two seats next to her. This leaves 5 places for him to sit. Then, this leaves 6 unique seats for the 6 remaining people, so there are 6! ways to sit them after Bob is sitting. So the answer is $ 5 \times 6! = \boxed{3600} $.
Freeflows