Ten people are to sit at a round table. Find the number of seating arrangements if the host and the hostess must always sit side by side.

Question

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 $8!\cdot 2!$ ?

Answer 1

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

Answer 2

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 $2\times8!$.

Answer 3

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