Find the number of rearrangements of the string 123456 in which none of the sequences 123, 321, 456, and 654 occur. My attempt:


Let $A_{1}$ denotes where 123 occurs, $A_{2}$ denotes where 321 occurs, $A_{3}$ denotes where 456 occurs, and $A_{4}$ denotes where 654 occurs. Also $|U|$=6!
$$\begin{aligned}|A_{1} &\cup A_{2} \cup A_{3} \cup A_{4}|\\
&= |A_{1}|+|A_{2}|+|A_{3}|+|A_{4}|\\
&\quad -(|A_{1}A_{2}|+|A_{1}A_{3}|+|A_{1}A_{4}|+|A_{2}A_{3}|+|A_{2}A_{4}|+|A_{3}A_{4}|)\\
&\quad +|A_{1}A_{2}A_{3}|+|A_{1}A_{2}A_{4}|+|A_{1}A_{3}A_{4}|+|A_{2}A_{3}A_{4}|\\
&\quad-|A_{1}A_{2}A_{3}A_{4}|\\
=& \ ? \end{aligned}$$ 
Then, 
$$\begin{aligned}|A_{1}'A_{2}'A_{3}'A_{4}'|& =|U|-|A_{1} \cup A_{2} \cup A_{3} \cup A_{4}|\\
&= 6!-? \end{aligned}$$


I am pretty sure this is ALL wrong. Please help me with this problem. Thank you!

Your approach is quite  ok.

We can enhance readability a bit by denoting the sets $A_1,A_2,A_3,A_4$    with
  \begin{align*}
A_{123},A_{321},A_{456},A_{654}
\end{align*} 
  instead. We  apply the IEP
  and obtain
  \begin{align*}
|A|&=6!-\left(|A_{123}|+|A_{321}|+|A_{456}|+|A_{654}|\right)\\
&\qquad+\left(|A_{123}\cap A_{456}|+|A_{123}\cap A_{654}|+|A_{321}\cap A_{456}|+|A_{321}\cap A_{654}|\right)
\end{align*}
We  do not  have to respect more terms, since $$A_{123}\cap A_{321}=A_{456}\cap A_{654}=\emptyset$$
Since $|A_{123}|=4!$ and $|A_{123}\cap A_{456}|=2!$ we get due to symmetry
  \begin{align*}
\color{blue}{|A|}&=6!-4\cdot4!+4\cdot 2!\\
&=720-96+8\\
&=\color{blue}{632}
\end{align*}

