Number of ways to arrange $A,A,A,B,C,C$ such that no $2$ consecutive letters are the same

Question

There is a question from my problem set that I am facing difficulty in solving.

It says to find the number of ways to arrange $A, A, A, B, C, C$ so that no $2$ consecutive letters are the same.

Some of the comments are suggesting this is a duplicate but in that question, the frequency of the letters was the same making it easier to solve by inclusion-exclusion principle. Not the case here, however.

My approach:

I tried to formulate cases where no $2$ consecutive letters would be the same. I tried using the gap method wherein I created a scenario like this,
$- A- A - A- $
Now we can see that there are $4$ spaces and three letters left to fill, and no restrictions, so we get $\binom{4}{3}\cdot 3! = 24$.
However, this is obviously not the only case where this is possible. I think this might be solvable with the inclusion-exclusion principle.

Approach 2:

If we take cases for the consecutive letters being the same, then we can perhaps subtract from the total number of cases, for example, we can take
$2 A$'s being consecutive and then $3A$'s and then $2B$'s and so on...
This is where I require help.
Any help would be appreciated.

Answer 1

First determine the order of the B, C, C, ignoring the As. There are $3$ possibilities for that.

Now you have, for example, -C-B-C- and you need to fill three of the four open slots with As. There are $4$ ways to do that -- you just have to choose which slot you don' fill.

The cases where this won't produce a valid sequence is if the two Cs were adjacent in the initial sequence and don't get an A between them. That is, we need to subtract two for exactly the sequences ABACCA amd ACCABA, and our final count is $$ 3\cdot 4 - 2 $$