In how many ways can eight people, denoted $A,B,C,D,E,F,G,H$ be seated about a square table that seats two people on each side?

Consider the complement: how many ways can they be seated such that $A$ and $B$ must sit next to each other? Well, imagine gluing the couple together as a single superperson $\boxed{AB}$. Note that we could also permute the two people within the superperson, so there are $2!$ ways to do this. Next, we must now permute the $7$ people, giving us $7!$ possibilities. But because a square can be rotated four times, we overcounted by a factor of $4$. Putting everything together and subtracting from the result from the first problem, we obtain: $$ \frac{8!}{4} - \frac{2!7!}{4} = \frac{8 \cdot 7! - 2 \cdot 7!}{4} = \frac{6 \cdot 7!}{4} = \frac{3 \cdot 7!}{2} = 7560 $$