Basic combinatorics question- what am I doing wrong? So the question is how many different stings of length 12 made of the characters AAABBBBCCCCC are there with no successive C's?
So my idea was to calculate the number of total possible strings with no restrictions-
$(12\ c\ 3) \times (9\ c\ 4)$
and then subtract the numbers of strings of length 11 where one of the characters is "CC", which gives me:
$(11\ c\ 3) \times (8\ c\ 4) \times (4\ c\ 3)$ options.  
But the latter is greater than the former.
What am I doing wrong here?
 A: As mentioned in the comments above, you have overcounted the number of "bad" strings.
Letting $X$ represent your "CC" character, the strings XCCCAAABBBB, CXCCAAABBBB, CCXCAAABBBB, CCCXAAABBBB all in reality refer to the single string CCCCCAAABBBB but with your calculations you have referred to it four separate times.  Similarly other strings are being referred to too many times as well.
For a correct approach, first consider the problem of arranging CCCCCXXXXXXX where no two C's are adjacent and then replace the string of X's with an arrangement of A's and B's.  For the first step, use stars-and-bars.
Letting $x_1$ be the number of $X$'s to the left of the first $C$, $x_2$ the number of $X$'s between the first and the second $C$, etc... on up to $x_6$ the number of $X$'s to the right of the final $C$, we have the system:
$\begin{cases}x_1+x_2+\dots+x_6=7\\0\leq x_1\\ 1\leq x_2\\ 1\leq x_3\\ \vdots\\ 0\leq x_6\end{cases}$
Via a change of variables, setting $y_i=x_i-1$ for each $i\in\{2,3,4,5\}$ and $y_i=x_i$ for $i\in\{1,6\}$ we have the new system:
$\begin{cases}y_1+y_2+\dots+y_6=3\\0\leq y_i~~\forall i\end{cases}$ which should be in a known form.

 There are $\binom{3+6-1}{6-1}=\binom{8}{5}$ such arrangements

Now, we replace the X's from left to right with an arrangement of three A's and four B's

 There are $\binom{7}{3}$ such arrangements

Multiplying we get the final total:

 $\binom{8}{5}\binom{7}{3}$

A: Treating $C$'s as boundaries, the five $C$'s splits the 12-character string into six regions:
$$\ldots C\ldots C\ldots C\ldots C\ldots C\ldots$$
We have to assign seven character slots into 6 regions. Note that the centre 4 regions by default has at least one character slots in it to separate the $C$'s. The remaining $3 = 7-4$ character slots with the $5$ $C$'s form a classical stars and bars problem:
How many ways are there to arrange $3$ slots (stars) and $5$ $C$'s (bars)? An example is
$$\_C\_CC\_CC = *|*||*||$$
There are $\binom83$ ways to arrange the three slots. This is exactly equal to the ways to arrange $7$ slots and $5$ $C$'s, so that no two $C$'s are adjacent. Continuing the example, adding the default slots:
$$\_C\_\_C\_\_\_C\_C$$
Independently, now there are seven slots regardless of the positions of $C$'s. For each of the position of $C$'s, there are $\binom 7 3$ ways to arrange $3$ $A$'s into $7$ slots. So the final answer is
$$\binom{8}{3}\binom 7 3$$
A: The problem is simply overcounting, since any string with $CCC$ will be counted twice, etc.
Working "the other way round" to make the background first, then add $C$s::
Calculate the number of arrangements of the $A$s and $B$s. There are $3$ $A$s, so the count of arrangements is $\binom 73 = 35$.
Now there are 8 "gaps" available which we can place one $C$ in ($6$ between letters and one at each end), and we have $5$ $C$s; so  $\binom 85 = 56$ choices. 
Multiply together for the result, $\binom 73\binom 85 = 35\cdot 56 = 1960$.
