14. Now there are 12 people sitting around a round table. If there are two people who do not want to sit in adjacent seats, how many different seating arrangements are there?
14. When not considering the restrictions in the problem, the circular permutation number for $n=12$ is 11! Assuming the two people who do not want to sit together are $A$ and $B$, when $A$ and $B$ sit together, it is equivalent to a circular permutation for $n=11$, with the permutation number being $10!$. And $A$ and $B$ being adjacent has two cases, i.e., $AB$ and $BA$. Therefore, the number of ways for $A$ and $B$ not to sit together is $11! - 2 \times 10! = 9 \times 10$!