What I'm thinking: Find total ways that the ten people can be seated, which is 10!.
Then I take that number and subtract the ways the these two people would be seated next to each other. I do this by treating these two people as a single space, which leaves the eight other students plus that space consisting of the two. This would mean 9!
Then, 10! - 9! = 3265920 ways for the ten people to be seated so that a certain to are not next to each other.
On
Another way of thinking is. You can freely position 9 people (including one of the special two), therefore $9!$. However the remaining one cannot sit next to the other special person, so only 8 possible positions. Overall $8 \times 9!$
On
The best way to solve this is take the 2 special people as a separate unit. Arrange the 8 other people in a row i.e. 8!. Since this is a linear arrangement, the number of gaps between the 8 already seated people will be 9. To select 2 gaps from 9 gaps we'll use 9C2 for A and B and multiply by 2! for arranging A and B (basically 9P2). So, the overall answer is 8!×9c2×2! Thank you.
Call the two special people $A$ and $B$. There are nine seats where the "left" one can sit and the right next to him. There are two such cases $AB$ and $BA$. For each of these conditions there are 8 remaining slots that can be filled by the remaining $8$ folk arbitrarily.
Thus: $9 \cdot 2 \cdot 8! = 725,760$
The total number of ways the chosen two people do not sit next to each other is $10! - 9 \cdot 2 \cdot 8! = 2,903,040$.