# count the ways of seating at the square table

In how many ways can eight people, denoted A, B, ..., H be seated about the square table shown where figs (a) and (b) are considered the same but are distinct from fig (c)? this part of problem is easy and the answer is $2\times 7!$ but I can't solve the rest of problem.

If two of the eight people say A and B do not get along well, how many different seatings are possible with A and B not sitting to each other.

I said if it means that 2 people can't seat on the same side of table then we can choose side and seat A and B on that side there is $8$ possibilities and then we have 6! ways for other ones and at the end we should divide answer by 4 because it's square table then we have $(2\times 7!) - (\frac{2\times 4 \times 6!}{4})$ and its equal to $8640$ but this solution count (c) as a valid seating. we can solve it in another way if (c) is in valid. If we count AB as one person then we have 7! seating possibilities that we count each one 4 times so the answer of problem is $(2\times 7!) - (\frac{2 \times 7!}{4})$ and it's equal to $7560$ but the solution manual said the answer is $7200$.

This problem is from first chapter of Discrete and Combinatorial Mathematics.

To consider A and B as one person is a little more complicated. The A and B on one side is equal to the A and B on the corner, so consider that case, the answer is just double the number substracted$$(2\times 7!)-2\times \frac{2\times 4 \times 6!}{4}=7200$$
To consider A and B as one person, can assume that AB take the first side, leaves 6 position for free choice + AB take the first corner, leaves 6 position for free choice. And double it because AB can be BA. so is also: $$(2\times 7!)-4\times 6!=7200$$
See where you made mistake is that $a$ can seat on any of the $8$ chairs now $b$ has only one option and they can be arranged in $2!$ ways all your other argument is right so now the ways where $a,b$ seat together is $\frac{8.2!.6!}{4}=2880$