Find the number of ways of arranging the letters Find the number of ways of arranging the letters $\text{AAAAA, BBB, CCC, D, EE &amp; F}$ 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.
Ignore the C's for the moment and arrange the remaining letters. There are $\frac{12!}{5!3!2!}=332640$ ways to do this.

Now consider the thirteen spaces between and beyond the letters &ndash; at most one C may be inserted into each space, so given an arrangement of the other letters there are $\binom{13}3=286$ ways to add C's.

The total number of admissible combinations is thus $332640×286=95135040$.
First arrange all the letters without $C$'s, i.e. $$\frac{12!}{5!3!1!2!1!}.$$
Now we have to fill inside the gaps between arranged letters.
As they given no two $C$'s come together, means $3$ $C$'s also doesn't come together.
The answer:
$$
\frac{12!}{5!3!1!2!1!}*\frac{P^{13}_3}{3!}.
$$
For example I have $3$ chocolates, that doesn't mean that I don't have $2$.