There are $\binom{6}{3} = 20$ sequences of three $A$'s and three $B$'s. Consider the ten sequences of three $A$'s and three $B$'s that begin with an $A$.
$\color{red}{AAABBB}$
$\color{green}{AABABB}$
$\color{green}{AABBAB}$
$\color{blue}{AABBBA}$
$\color{green}{ABAABB}$
$\color{cyan}{ABABAB}$
$\color{magenta}{ABABBA}$
$\color{green}{ABBAAB}$
$\color{magenta}{ABBABA}$
$\color{blue}{ABBBAA}$
No matter how we place $C$'s in the sequence $\color{red}{AAABBB}$, at least two consecutive letters will be the same.
There is only one way to place the $C$'s in the sequences $\color{blue}{AABBBA}$ and $\color{blue}{ABBBAA}$ since we are forced to place a $C$ between each pair of consecutive $B$'s and the pair of consecutive $A$'s.
The number of ways we can fill three of the seven spaces (the beginning, the end, and the five spaces between consecutive letters) in the sequence $\color{cyan}{ABABAB}$ is $\binom{7}{3}$.
We must place a $C$ between the pair of consecutive $B$'s in the sequences $\color{magenta}{ABABBA}$ and $\color{magenta}{ABBABA}$, which leaves us $\binom{6}{2}$ ways to insert the two remaining $C$'s in the six remaining spaces.
We must place one $C$ between the pair of consecutive $A$'s and another $C$ between the pair of consecutive $B$'s in the sequences $\color{green}{AABABB}$, $\color{green}{AABBAB}$, $\color{green}{ABAABB}$, $\color{green}{ABBAAB}$, leaving five spaces in which to place the remaining $C$.
Hence, the number of sequences of three $A$'s, three $B$'s, and three $C$'s that begin with $A$ that do not contain consecutive letters that are the same is
$$1 \cdot 0 + 2 \cdot 1 + 2 \cdot \binom{6}{2} + 4 \cdot \binom{5}{1} + 1 \cdot \binom{7}{3} = 0 + 2 + 30 + 20 + 35 = 87$$
By symmetry, there are also $87$ such sequences that begin with a $B$. Hence, the number of sequences of three $A$'s, three $B$'s, and three $C$'s that do not contain consecutive letters that are the same is $2 \cdot 87 = 174$.
The original poster wanted a solution with generating functions. So here is my take on that:
I would use ordinary generating function instead of exponential one but still write:
$ A= \frac{x^{4}}{24} + \frac{x^{3}}{6} + \frac{x^{2}}{2} $
Where the motivation is: Since all the As are indistinguishable we have to divide by the number of permutations within those As. If e.g. $x³$ is choosen divide by 6.
The above kind of looks like an exponential generating function but the divisions are just to account for removing the permutation within the As.
$10! \cdot A^4 = \frac{175 x^{16}}{16} + 175 x^{15} + \ldots + 109200 x^{11} + 226800 x^{10} + 302400 x^{9} + 226800 x^{8}$
Where you can read off the solution at coefficient $x^{10}$
The original poster asked on how to find this. In the end generating functions only transform one combinatorial problem into another. The nice thing is that with that the one expressed in terms of generating functions can be handed of the a CAS system (I used sympy here).
Update
The above is a bit of a hack as the multiplication by $10!$ only makes it correct for the 10 digit case. If the the function is indeed seen as an exponential generating functions we can read of all digits
$ \frac{x^{16}}{331776} + \frac{x^{15}}{20736} + \frac{x^{14}}{2304} + \frac{13 x^{13}}{5184} + \frac{107 x^{12}}{10368} + \frac{13 x^{11}}{432} + \frac{x^{10}}{16} + \frac{x^{9}}{12} + \frac{x^{8}}{16} $
Or:
$ A^4=63063000 \frac{x^{16}}{16!} + 63063000 \frac{x^{15}}{15!} + 37837800 \frac{x^{14}}{14!} + 15615600 \frac{x^{13}}{13!} + 4943400 \frac{x^{12}}{12!} + 1201200 \frac{x^{11}}{11!} + 226800 \frac{x^{10}}{10!} + 30240 \frac{x^{9}}{9!} + 2520 \frac{x^{8}}{8!} $
Best Answer
First place the $A$'s in $n!$ ways. This can be seen through the lens of permutation matrices.
Next, given some placement of the $A$'s, we wish to place the $B$'s in such a way as to never reuse the same positions. Through a relabeling of indices, this is effectively asking for a derangement and can be done in $!n$ ways where $!n$ is the subfactorial counting the number of derangements of $[n]$.
We have a final total then:
$$n!\cdot !n$$