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

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 $$B$$.

Also

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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:

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

But the answer is $$5760$$

• 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
$$\frac{8 \cdot 6 \cdot 6!}{4} = 8640$$
• @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. Commented Oct 2, 2020 at 8:23