The question is:
Atticus has three comedies, two dramas and an action movie. He decides that he wants to watch two, and only two, comedies in a row. In how many ways can Atticus watch his six movies?
The given answer is $432$.
My attempts:
Attempt 1:
The number of ways in which he can watch at least two comedies in a row can be given by fixing two of them together:
$C_1\,X_1\,(C_2\,C_3)\,X_2\,X_3$ ($5!\cdot3!$)
Then we subtract the number of ways he watches all three comedies in a row:
$X_1\,(C_1\,C_2\,C_3)\,X_2\,X_3$ ($4!\cdot3!$)
$5!\cdot3!-4!\cdot3!=576\neq 432$
Attempt 2:
There are $6!$ ways he can watch movies without any restriction.
There are $4!\cdot3!$ ways he can watch all three comedies in a row.
The number of ways he watches exactly one comedy in a row is:
$C_1\,X_1\,C_2\,X_2\,C_3\,X_3$ ($3!\cdot3!$)
$X_1\,C_1\,X_2\,C_2\,X_3\,C_3$ ($3!\cdot3!$)
So to find the ways in which he watches exactly two comedies in a row we can take the total without restriction and subtract the number of ways he can watch, exactly, either one or three comedies in a row.
$6!-2\cdot (3!)^2 -4!\cdot3! = 504 \neq 432$
What cases am I missing?
Arrange the non-comedy first, there are $3$ of them: $\color{blue}{3!}$. Denote them as $N$.
$$XNXNXNX$$
where $X$ are slots to place the comedies, let me tie up two comedies together. Hence I have two block of comedies, one consists of one comedy and one consists of two. number of ways to form the block: $\color{blue}{\binom{3}{1}}$
Each $X$ can take in at most one block. Also, remember that we can change the order within the block: $$2! \cdot \binom{4}{2}\cdot 2$$
Hence in total
$$3! \cdot 3 \cdot 2! \cdot \binom{4}{2}\cdot 2 = 6(3)(2)(6)(2)=432$$
Edit:
Remark for attempt $1$:
Have you remove over counting like $(C_1C_2)C_3$ and $C_1(C_2C_3)$?
Remark for attempt $2$:
Have you considered cases like $C_1X_1X_2C_2X_3C_3$?
Number of ways to watch exactly one movie can be counted by arranging $X$ first $3!$ and then prepare slots to insert exactly one comedy, there are $4$ slots. $\binom{4}{3}\cdot 3!$
Hence $$6!-4(3!)^2-4!3!=432$$