What are the chances you and your friend will NOT be seated together in a row of 12 seats? The answer is 5/6 but I think its 10/11 for some reason.
Why I think it's 10/11 is because after my friend is seated there are only 11 available, meaning there is 1 seat out of 11 that i can sit in so I am beside them.
Can anyone explain how to do this problem, using the combination formula perspective if possible.
 A: The total number of possible arrangements equals $12!$. The number of arrangements in which you and your friend are sitting together equals $11!×2!$, since we consider the duo as block, then we have $11!$ possible arrangements along with $2!$ internal arrangements of the friends. Hence the total number of arrangements in which you and your friend will be seated together equals:
$$\frac{11!×2}{12!} = \frac{1}{6}$$
Therefore the chances that you and your friend will not be seated together equals:
$$1 - \frac{1}{6} = \frac{5}{6}$$
A: There is not only one seat that is next to your friend after they are seated. Sometimes there are two, depending on whether they are on the end of the row or not.
It's easier to compute the chances you are seated together. It's the number of seating arrangements in which you are next to your friend, divided by the total number of seating arrangements. The total number of seating arrangements is $12!$ ($12$ choices for who goes in the first seat, $11$ choices for who goes in the second, and so on). 
To count the number of seating arrangements in which you are next to your friend, imagine the two of you as a single unit that cannot be separated. We want to arrange the $10$ other people plus the pair of you, which we are counting as one unit that cannot be separated. There are $11!$ ways to do so ($11$ choices for which person/pair goes first, $10$ for which goes second, and so on). But for every such arrangement, there are $2$ ways to arrange you and your friend relative to one another. So there are $2\cdot 11!$ seating arrangements in which you are next to your friend.
So the probability you are next to each other is $\frac{2\cdot 11!}{12!}=\frac{2}{12}=\frac{1}{6}$, and thus the probability that you are not next to each other is $\frac{5}{6}$.
Edit: I did not see that you wanted to use combinations rather than permutations. John Douma's comment is a nice easy solution using combinations.
