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.

While I prefer the approach explained by Alaleh A and Parcly Taxel, here is how you can solve the problem using the Inclusion-Exclusion Principle.
The number of distinguishable arrangements of $5$ A's, $3$ B's, $3$ C's, $1$ D, $2$ E's, and $1$ F is 
$$\frac{15!}{5!3!3!1!2!1!}$$
as you found.
From these, we must subtract those arrangements in which a pair of C's are adjacent.
A pair of C's are adjacent:  We have $14$ objects to arrange, $5$ A's, $3$ B's, $1$ CC, $1$ C, $1$ D, $2$ E's, and $1$ F.  They can be arranged in 
$$\frac{14!}{5!3!1!1!1!2!1!}$$
ways.
However, if we subtract $\frac{14!}{5!3!1!1!1!2!1!}$ from the total, we will have subtracted too much since we will have subtracted each arrangement with two pairs of adjacent C's twice (such arrangements have three consecutive C's), once when we designated the first two C's as the adjacent pair and once when we designated the last two C's as the adjacent pair.  Since we only want to subtract such arrangements once, we must add them back.
Two pairs of adjacent C's:  As mentioned above, this means the three C's are consecutive.  Hence, we have $13$ objects to arrange, $5$ A's, $3$ B's, $1$ CCC, $1$ D, $2$ E's, and $1$ F.  The number of such arrangements is 
$$\frac{13!}{5!3!1!1!2!1!}$$
as you found.
Hence, the number of arrangements of $5$ A's, $3$ B's, $3$ C's, $1$ D, $2$ E's, and $1$ F in which no two C's are consecutive is 
$$\frac{15!}{5!3!3!1!2!1!} - \frac{14!}{5!3!1!1!2!1!} + \frac{13!}{5!3!1!1!2!1!}$$
