Need help with combinatorics question(probably cyclical permutation) A human invites 6 of his friends to a meeting. In how many different arrangements they along with the human's wife can sit at a round table if the hosts and the wife always sit together?
Is this a cyclical permutation problem? Please explain.
 A: First of all you place wife and husband at the table (you can do this in $8\cdot 2$ ways, since you can place a pair at any place and for each choice you can switch wife and husband). Then you place left guests in $6!$ ways. So there are $8\cdot2\cdot6!$ arrangements.
A: It is probably intended that any rotation of the people is considered to produce the"same" arrangement, meaning that yes, this is intended to be a cyclic permutation problem. 
That is not the only possible interpretation, since the host might like to be nearest to the kitchen. And the views from the chairs are different. 
To solve, since we can rotate the people without changing the arrangement, let one of the chairs be a throne, and let the host's spouse sit there. Then the host can be placed in $2$ ways. And for each of these ways, the $6$ guests $\dots$. 
A: You have 8 people. We know that the host and his wife must sit together. So we sit the host first: we have 8 choices, and, for each choice, we can sit his wife either on his right or his left; thus we have $8\times2$ ways. Now, once you have fixed the position of the host and his wife, you can sit the 6 guests as you like, i.e. in $6!$ ways. So in total we have $2\times8\times6!=11520$ arrangements.
