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In how many ways the $8$ people $A,B,C,D,E,F,G,H$ can be arranged around a square table assuming $A$ should not be seated in front of $B$.

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The rightmost one is the same as the middle one, however they two are different from the leftmost one.

The number of arrangements that $8$ people can sit around such a table is $2\cdot7!$ On the other hand for each one of the sides one of the two cases happens:

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The number of such arrangements is $2\cdot6!$, so the desired answer is $2\cdot7!- 2\cdot6!=12\cdot6!=8640$

But the answer is $5760$

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Your answer is correct. Look at it this way:

  • First, seat $A$ in $8$ possible ways
  • Then, seat $B$ in $6$ possible ways ($B$ not directly in front of $A$)
  • Finally, seat the remaining people in $6!$ possible ways

Accounting for the four rotations of the table, we find:

$$\frac{8 \cdot 6 \cdot 6!}{4} = 8640$$

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    $\begingroup$ @45465 I found the 5th edition, which mentions that $A$ and $B$ cannot be seated next to one another. Here, the answer is $\frac{8 \cdot 5 \cdot 6!}{4} = 7200$ (verified solution here). Looking back at it, $5760$ would be the number of ways in which $A$ and $B$ are not seated next to each other or in front of each other. $\endgroup$
    – jvdhooft
    Commented Oct 2, 2020 at 8:23

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