In how many ways can ten people be seated in a row so that a certain two of them are not next to each other? 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. 

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 $10! - 9 \cdot 2 \cdot 8! = 2,903,040$.
