Although cumbersome, Inclusion-Exclusion eliminates much of the need for creativity. See this article for an
introduction to Inclusion-Exclusion.
Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
Using the syntax in the second link:
let $S$ denote the set of all possible permutations, without any regard for which letters are adjacent.
let $S_1$ denote the subset of $S$ that contains all of the permutations where the two A's are adjacent.
let $S_2, S_3$ and $S_4$ be similarly defined for the B's, C's, and D's, respectively.
Then, the desired computation is
$$|S| - |S_1 \cup S_2 \cup S_3 \cup S_4|.$$
Continuing to use the syntax of the 2nd link:
let $T_0$ denote $|S|.$
let $T_1$ denote the $~\displaystyle \binom{4}{1}$ terms given by $\displaystyle \sum_{1 \leq i_1 \leq 4} |S_{i_1}|.$
let $T_2$ denote the $~\displaystyle \binom{4}{2}$ terms given by $\displaystyle \sum_{1 \leq i_1 < i_2 \leq 4} |S_{i_1} \cap S_{i_2}|.$
let $T_3$ denote the $~\displaystyle \binom{4}{3}$ terms given by $\displaystyle \sum_{1 \leq i_1 < i_2 < i_3 \leq 4} |S_{i_1} \cap S_{i_2} \cap S_{i_3}|.$
let $T_4$ denote the term given by $|S_1 \cap S_2 \cap S_3 \cap S_4|.$
Then, in accordance with Inclusion-Exclusion theory, the desired computation is
$$\sum_{k=0}^4 \left[(-1)^k T_k\right]. \tag1 $$
In some of the analysis below, Stars and Bars theory will also be used.
See this article and
this one.
$\underline{\text{Computation of} ~T_0}$
Since there are $(4)$ pairs of identical letters to permute,
$$\displaystyle T_0 = \frac{8!}{(2!)^4} = 2520. \tag2 $$
$\underline{\text{Computation of} ~T_1}$
First, compute $|S_1|$.
The two A's can be in any one of $(7)$ positions.
Assuming that the two A's are in the first two positions, there are $(6)$ remaining letters to permute.
Therefore, $\displaystyle ~|S_1| = 7 \times \frac{6!}{(2!)^3} = 630.$
By reasons of symmetry, $|S_1| = |S_2| = |S_3| = |S_4|.$
Therefore,
$$T_1 = 4 \times 630 = 2520. \tag2 $$
$\underline{\text{Computation of} ~T_2}$
First, compute $|S_1 \cap S_2|$.
Either the A's come first, or the B's come first.
Therefore, you start with a factor of $(2)$, and then assume that the $A's$ come first.
You can regard the A's and B's as forming two fused units, with the A's unit coming first. Then, there are potentially $(3)$ undeclared regions: before the A's, between the A's and B's, and after the B's.
These $(3)$ regions must account for the $(4)$ remaining unused letters. The number of ways that the A's and B's can be positioned is equal to the number of non-negative integer solutions to
$x_1 + x_2 + x_3 = 4 ~: ~\displaystyle \binom{6}{2} = 15~$ solutions.
Then, for each one of the $(15)$ solutions, there are $~\displaystyle \frac{4!}{(2!)^2} = 6$ ways that the C's and D's can be permuted.
Therefore,
$\displaystyle |S_1 \cap S_2| = 2 \times 15 \times 6 = 180.$
By reasons of symmetry, the other $\displaystyle \left[\binom{4}{2} - 1\right]$ terms involved in the computation of $T_2$ are identical.
Therefore,
$$T_2 = 6 \times 180 = 1080. \tag3 $$
$\underline{\text{Computation of} ~T_3}$
First, compute $|S_1 \cap S_2 \cap S_3|$.
The A's, B's, and C's can be permuted in $(3!) = 6$ ways.
Therefore, you start with a factor of $(6)$, and then assume that the $A's$ come first, then the B's, and then the C's.
Similar to the analysis in the previous section, the corresponding number of arrangements is equal to the number of non-negative solutions to
$\displaystyle x_1 + x_2 + x_3 + x_4 = 2 ~: ~\binom{5}{3} = 10~$ solutions.
Here, once the A's, B's, and C's are set, the two (identical) D's can only be positioned in one way.
Therefore,
$\displaystyle |S_1 \cap S_2 \cap S_3| = 6 \times 10 = 60.$
Similar to the previous section, by reasons of symmetry, the other $\displaystyle \left[\binom{4}{3} - 1\right]$ terms involved in the computation of $T_3$ are identical.
Therefore,
$$T_3 = 4 \times 60 = 240. \tag4 $$
$\underline{\text{Computation of} ~T_4}$
The $(4)$ pairs can be permuted in $(4!)$ ways. For any given ordering of the letters, there is no wiggle room here. That is, the first pair must go in positions 1 and 2, and so forth.
Therefore,
$$T_4 = 4! = 24. \tag4 $$
$\underline{\text{Final Computation}}$
$$(T_0 + T_2 + T_4) - (T_1 + T_3) = (2520 + 1080 + 24) - (2520 + 240) = 864.$$
