Find the number of ways of arranging the letters 
Find the number of ways of arranging the letters $\text{AAAAA, BBB, CCC, D, EE & F}$ in a row if no two $\text{C's}$ are together?

My Attempt:
Well, I should I arrive at the answer if I subtract the cases where 3 $\text{C's}$ and 2 $\text{C's}$ appear together from the total.
Total possibilities $= \frac{15!}{5!3!3!2!}$
Total possibilities where 3 $\text{C's}$ appear $=\frac{13!}{5!3!2!}$
However, I am not able to find the possibilities for 2 $\text{C's}$ being together and get to the answer.
Any help would be appreciated.

Let $n_1$ be all letters except C ($=12$)and $n_2$ be Cs ($=3$).  I am going to generalize the solution in such a manner that you have no two C's together (p=0), two C's together (p=1) and three C's together (p=2)
Each pair must have a left-hand member, which we can choose from any of the Cs except the last: $\tbinom{n_2-1}{p}$. This will create $n_2-p$ blocks of Cs, which we can distribute into the $n_1+1$ slots between and around the non-Cs: $\tbinom{n_1+1}{n_2-p}$. Thus the number of arrangemenets with exactly $p$ pairs is $\tbinom{n_2-1}{p}\tbinom{n_1+1}{n_2-p}.\frac{n_1!}{5!.3!.2!}$
For p = 0 in the question
$$\tbinom{3-1}{0}\tbinom{12+1}{3-0}.\frac{12!}{5!.3!.2!}$$
$$\tbinom{2}{0}\tbinom{13}{3}.\frac{12!}{5!.3!.2!}$$
