# Permutations and Combinations Question?

Mr and Mrs Jones and 6 guests sit around the dinner table. In how many ways can they be arranged if the two host are separated?

The answer says 3600, but I could never get that.

My working out was 7! - 2!5! (total number of permutations minus the permutations of the hosts sitting together)

Any help would be appreciated.

You can seat Mrs Jones in any of $8$ places. Once she’s seated, there are $3$ places that are not available to Mr Jones: the place where she’s seated, and the two adjacent places. Thus, there are $5$ choices for his seat. That leaves $6$ people who can fill the remaining $6$ seats in any of $6!$ possible orders, for a total of $8\cdot5\cdot6!=28,800$ possible arrangements if the seats are the table are individually identified.
If we care only about who sits next to whom, then we don’t care which of the $8$ seats Mrs Jones takes: all we care about is where the others sit in relation to her. That reduces the number of choices to $5\cdot6!=3600$ and must therefore be the intended interpretation.
The eight persons can sit around the round table in $(8-1)!$ distinct ways. No of arrangements where the hosts sit together is equal to the number of distinct ways & person can sit around the round table (think of the couple as a single person), i.e. $(7-1)!$, multiplied by 2 (Mr. Jones can sit to the right of Mrs. Jones and Mrs. Jones can sit to the right of Mr. Jones). So total number of ways is $$7!-2\times6!=3600$$