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.

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:
$$1 - \frac{1}{6} = \frac{5}{6}$$
