Alternatives
Seat Alex in a random chair. There are now $7$ seats remaining, of which $2$ are next to Alex. If we seat Bob randomly, then with probability $2/7$ he sits next to Alex. So, in $2/7$ of all possible cases, Alex sits next to Bob. There are $8!$ different ways in which you can seat 8 persons, so the number of ways how you can seat $8$ people at a round table so that Alex and Bob sit next to each other is
\begin{equation}
\frac{2}{7} \cdot 8! = 11520.
\end{equation}
Label the seats $1,2,\ldots,8$. There are $8$ groups of $2$ seats that are next to each other: $\{(1,2),(2,3),\ldots,(7,8),(8,1)\}$. Within each group, there are $2!$ ways to Alex and Bob. Outside each group there are $6!$ ways to seat the remaining $6$ persons. So, the number of ways how you can seat $8$ people at a round table so that Alex and Bob sit next to each other is
\begin{equation}
8 \cdot 2! \cdot 6! = 11520.
\end{equation}
Something which would affect the answer is if "one seat away" is actually shorthand for "at least one seat away" (e.g. they'll fight if they're next to each other).
– Joffan Jun 02 '16 at 14:40