If 8 people consisting of 4 couples are randomly arranged in a row, find the probability that no person is next to their partner. I understand how to arrive at the correct answer using the inclusion-exclusion principle. The purpose of this post is so I can understand why the reasoning I first used is incorrect.
My reasoning: there are T = 8! total ways to arrange the 8 people. Define $A_i$ to be the number of ways to arrange 8 people in a line such that exactly i of the 4 couples are seated next to each other. Since T = $\sum_{i=0}^4 A_i$, I attempted to find $A_0$ by computing $A_1$ through $A_4$ and subtracting from T.
$A_4$: Arrange 4 couples in 4! ways, and for each couple, 2 arrangements. So $2^4$ * 4! = 384.
$A_3$: Pick 3 of the 4 couples to be seated together, each of these couples can be arranged in 2 ways, and there are 5 entities (3 couples and 2 individuals) so 5! ways to arrange the 5 entities, thus ${4 \choose 3}$ * $2^3$ * 5!. However, this has counted the ways counted in $A_4$. So $A_3$ = ${4 \choose 3}$ * $2^3$ * 5! - $A_4$ = 3840 - 384 = 3456.
$A_2$:  Pick 2 of the 4 couples to be seated together, each of these couples can be arranged in 2 ways, and there are 6 entities (2 couples and 4 individuals) so 6! ways to arrange the 6 entities, thus ${4 \choose 2}$ * $2^2$ * 6!. But this has counted the ways counted in $A_3$ and $A_4$, so $A_2$ = ${4 \choose 2}$ * $2^2$ * 6! - $A_3$ - $A_4$ = 17280 - 3456 - 384 = 13440.
$A_1$:  Pick 1 of the 4 couples to be seated together, the couple can be arranged in 2 ways, and there are 7 entities (1 couple and 6 individuals) so 7! ways to arrange the 7 entities, thus ${4 \choose 1}$ * $2$ * $7!$. But this has counted the ways counted in $A_2$ and $A_3$ and $A_4$, so $A_1$ = ${4 \choose 1}$ * $2$ * $7!$ - $A_2$ - $A_3$ - $A_4$ = 40320 - 13440 - 3456 - 384 = 23040.
Now, $\sum_{i=1}^4 A_i$ = 40320 = T, implies $A_0$ = 0 which is obviously not true.
Where have I gone wrong?
 A: First of all, I agree with the comment of Henry.  This is (probably) not a duplicate question, because the original poster is specifically asking where he went wrong.
The entire approach is creative and flawed.  As a specific example, your computation of $A_3$ is wrong, because you assumed that the single deduction of $(384)$ was sufficient.
Consider the following four subsets represented by the computation of $A_3$:

*

*$M_1W_1, M_2W_2, M_3W_3$ are all together and $M_4W_4$ may or may not be together.


*$M_1W_1, M_2W_2, M_4W_4$ are all together and $M_3W_3$ may or may not be together.


*$M_1W_1, M_3W_3, M_4W_4$ are all together and $M_2W_2$ may or may not be together.


*$M_2W_2, M_3W_3, M_4W_4$ are all together and $M_1W_1$ may or may not be together.
So, in each of the $4$ scenarios, the situation where all $4$ couples are together is being counted.  So, you have to deduct $(384)$ four times, rather than only once.
I haven't examined the $A_2 \leftrightarrow A_3$ or $A_1 \leftrightarrow A_2$ scenarios, but I assume that there is a similar flaw there.
I think that your direct approach is workable, but necessarily convoluted.  If you want to pursue the direct approach, then I suggest that you use the computation from Inclusion-Exclusion as a guide, refining your direct approach to remove all of the flaws.
For what it's worth, perhaps an easier strategy of utilizing the direct approach is to compute $A_3$ in such a way that when $M_4,W_4$ is the odd couple out, you immediately prevent $M_4,W_4$ from being seated together.
