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The formula for circular permutation is $(n-1)!$, so if the two would sit next to each other, I'm not really sure of the computation. Would it be $(10-1)! \cdot 2!$ or $8!\cdot 2!$ ?

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Answer is $2 \times 8! $

As was mentioned by MysteryGuy number of all combinations is $2 \times 10 \times 8! $ (10 places for host, 2 for hostess, and 8! number of permutation for eight remaining guests).

But you mention circular permutation so I guess multiplier $10$ is excessive, because you can place host at some specific place and obtain all another permutations for all remaining hosts places by translation

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  • $\begingroup$ mmmm yes very correct $\endgroup$
    – mid
    Commented Apr 1, 2018 at 20:45
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Starting from where the host sits, the hostess can sit on their left or on their right. The remaining 8 people can sit in the remaining 8 chairs in any order. So the total number of seating arrangements (if we count rotations of an arrangement as being the same arrangement) is $2\times8!$.

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  • $\begingroup$ You forget to multiply by $10$ as there are 10 possibilities for the host to sit :) $\endgroup$
    – MysteryGuy
    Commented Mar 29, 2018 at 14:42
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    $\begingroup$ I would say this is probably the answer they are looking for. I think with the specification of "round table" you only care about the relative positions of people at the table. Therefore the other answer has an extra factor of $10$. $\endgroup$
    – wgrenard
    Commented Mar 29, 2018 at 14:52
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    $\begingroup$ @MysteryGuy - the point of specifying a circular table is that rotations don't count as distinct arrangements. So 0123456789 is the same seating arrangement as 1234567890 etc. You don't need the extra factor of 10 since the host, in effect, always sits in the same place. $\endgroup$
    – gandalf61
    Commented Mar 29, 2018 at 15:18
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First of all, you have $10$ possibilities for the host. Once he has a place, there are two remaining places for his wife. And finally $8! $ for the others.

So the answer is $2 \cdot 10 \cdot 8! $

Consider thinking logically rather than memorizing formula

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    $\begingroup$ Your answer is correct if you consider all rotations of a given arrangement as being distinct. I would guess that this is not what they are looking for, however. Since they specify that this is a round table I think they want you to identify all $10$ rotations of a given arrangement as the same arrangement. In this case, your answer over counts by a factor of $10$ $\endgroup$
    – wgrenard
    Commented Mar 29, 2018 at 14:56

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