The problem is as follows - there are three couples and six chairs in a row. The six individuals are seated at random. What is the chance that at least one couple will be seated together?
Here's my thinking - I treat a single couple as a unit. Since a couple takes up 2 seats, there are 5 positions the couple can fill. Since the orientation of the couple (MF or FM) doesn't matter, that doubles the number of permutations that fulfill the criteria. There are 3 couples that could do this, so I triple the number of permutations. That gives me 30 different ways that the set could be assembled within the criteria. However, each one of these has 4! permutations within it for different ways the seats around the couples could be filled. Therefore I have 4!30 ways that the couples could be seated next to each other. There are 6! ways for the whole arrangement to be set up. Divided out, this is equal to 1.
Obviously I made a mistake at some point, but I don't know where. If somebody could help explain my mistake I would greatly appreciate it.
The most obvious flaw in your reasoning is that it overcounts arrangements for which more than one couple sits together. For simplicity of notation, let the couples be $\{1,2\}$, $\{3,4\}$, $\{5,6\}$, where the females are the odd values and the males are the even values. Then the arrangement $(1,2,3,4,5,6)$ is counted numerous times depending on whether you count $(1,2)$, $(3,4)$, or $(5,6)$ as the "unit" that sits together.
Here is a hint on how to count the desired arrangements properly. Count instead the complementary event where no couples sit together.
Inclusion-exclusion is doable but a strict enumeration is also possible and serves as a good exercise. Think about it this way. There are obviously $6$ ways to choose the person sitting in the first seat. Now, there are only $4$ ways to choose the person sitting in the second such that no couples sit together; for if, say, person $1$ sits in the first seat, then person $2$ cannot sit in the next. Thus the first two seats are occupied by members of two distinct couples.
Now there are two cases for the third seat: either the third seat is occupied by the partner of the first seat's occupant, or the third seat is occupied by a member of the couple for whom neither has been seated. In the first case, there is only one unique choice for the third seat; then the fourth seat can be taken up in only two ways because if the fourth seat is taken by the partner of the second seat's occupant, then the remaining two seats must be taken by the unseated couple, forcing them together. So in the first case, there are $$6 \times 4 \times 1 \times 2 \times 1 \times 1 = 48$$ possibilities. In the second case, there are $2$ ways to choose the occupant of the third seat, as drawn from the couple who have not yet been seated. Once this occurs, there are again $2$ ways to draw from the remaining three people for the fourth seat; and finally $2$ ways to select the fifth person, for a total of $$6 \times 4 \times 2 \times 2 \times 2 \times 1 = 192$$ possibilities. Thus there are a total of $240$ outcomes in which no couples sit together.