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.)
I need to count the number of circular permutations of 8 people, but exclude those where Alice and Bob are adjacent. So the total number of ways to seat 8 people around a table is 7!. But how many of those have Alice and Bob next to each other? I can think of Alice and Bob as a single unit, and then there are 7 units to arrange around the table. That gives me 6! circular permutations. But Alice and Bob can swap places within their unit, so I need to multiply by 2. That gives me 2 * 6! circular permutations where Alice and Bob are adjacent. Therefore, the number of ways to seat 8 people around a table where Alice and Bob are not adjacent is 7! - 2 * 6!. I can simplify this expression by factoring out 6!. I get 6! * (7 - 2) = 6! * 5. I can evaluate this using a calculator or by hand. I get 6! * 5 = 720 * 5 = 3600. 