First, judging from your work, you misstated the question: it appears that you meant to ask for the probability that no husband sits next to his wife. It also appears that the seats are arranged in a row, not a circle. I will make these assumptions.
Your first three factors, $8\cdot6\cdot5$, in the numerator are fine. After that, though, you have to split the calculation into cases. Suppose that the first three people, in order, are A, B, and C. If A and C are a couple, then any of the remaining $5$ people can sit in the fourth seat. Your calculation, with a factor of $4$, is correct only if C’s spouse has not already been seated, i.e., C and A are not a couple. And the cases just proliferate after that, which is why you’re better off working with the complement.
We have persons $a,b,c,d,e,f$ with persons $a$ and $b$ being married and needing to sit next to one another (and no other married couples needing to be sat together). We want to count the number of ways these six people can be sat along a row (with a distinct start and end)
For the moment, denote the married couple $a,b$ as a single entity: $X$. How many ways do we have to arrange $X,c,d,e,f$ in a row? (e.g. one such outcome could look like $cXefd$)
$5! = 120$
Now that we have arranged the "five", replace $X$ with $a$ and $b$ again, but decide how to arrange $a$ and $b$ when putting it back in. (e.g. from $cXefd$ we now make it either $cabefd$ or $cbaefd$)
either $a$ is on the left or on the right: two possibilities.
By multiplication principle, the total number of arrangements is the multiplication of the number of choices at each of these two steps for a total of:
$5!\cdot 2 = 120\cdot 2 = 240$
If instead there are multiple married couples, each of which need to sit next to eachother then do similarly. Suppose $a$ and $b$ are married and need to sit next to eachother, $c$ and $d$ are married and need to sit next to eachother, and $e$ and $f$ are married and need to sit next to eachother.
For the moment, treat $a$ and $b$ as one collective entity: $X$. treat $c$ and $d$ as one collective entity: $Y$. treat $e$ and $f$ as one collective entity: $Z$.
How many ways are there to arrange $X,Y,Z$? (e.g. $YXZ$)
$3!$
Replacing $X$ with $a$ and $b$, how many ways can you do that? (e.g. $YbaZ$)
$2$
Replacing $Y$ with $c$ and $d$, how many ways can you do that? (e.g. $cdbaZ$)
$2$
Similarly, replace $Z$ with $e$ and $f$, in how many ways? (e.g. $cdbaef$)
$2$
By multiplication principle, multiply the number of ways of accomplishing each step for a total of:
$3! \cdot 2\cdot 2\cdot 2 = 8\cdot 6 = 48$
Given your statement in your post that this was the answer you expected, it appears that this was the interpretation and that you had neglected to mention that there are multiple married couples, all of which must sit near one another.
Best Answer
Alice has to be seated somewhere. After Alice is seated, there are 14 remaining seats where Bob could sit. 12 of those remaining seats are not next to Alice.
Therefore, the probability that Bob and Alice are not next to each other is $\frac {12} {14} = \frac 6 7$