How many different arrangements are there problem How many different arrangements are there of all the nine letters A, A, A, B, B, B, C, C, C in a row if no two of the same letters are adjacent?
First I tried to find how many ways to arrange so at least two similar letters are adjacent (the complementary) then subtract from total ways without restriction. I tried to do this via extended addition rule (i.e. with the three circle venn diagram) and now I'm confused how to calculate each case. 
My attempt so far: 
let a be set of 'two A's adjacent to each other'
let b be set of 'two B's adjacent to each other'
let c be set of 'two C's adjacent to each other'
I need to find |complement of a U b U c| (Let Z be universal set, no restriction)
= |Z| - |a| - |b| - |c| + |ab| + |bc| + |ca| - |abc|
I know |a| = |b| = |c| and |ab| = |bc| = |ca| 
therefore |complement of a U b U c| = |Z| - 3|a| + 3|ab| -|abc|.
|Z| = 9!/3!3!3!, but I'm not sure how to compute |a| or |ab| or |abc|
 A: There are $\binom{6}{3} = 20$ sequences of three $A$'s and three $B$'s.  Consider the ten sequences of three $A$'s and three $B$'s that begin with an $A$.
$\color{red}{AAABBB}$
$\color{green}{AABABB}$
$\color{green}{AABBAB}$
$\color{blue}{AABBBA}$
$\color{green}{ABAABB}$
$\color{cyan}{ABABAB}$
$\color{magenta}{ABABBA}$
$\color{green}{ABBAAB}$
$\color{magenta}{ABBABA}$
$\color{blue}{ABBBAA}$
No matter how we place $C$'s in the sequence $\color{red}{AAABBB}$, at least two consecutive letters will be the same.
There is only one way to place the $C$'s in the sequences $\color{blue}{AABBBA}$ and $\color{blue}{ABBBAA}$ since we are forced to place a $C$ between each pair of consecutive $B$'s and the pair of consecutive $A$'s.  
The number of ways we can fill three of the seven spaces (the beginning, the end, and the five spaces between consecutive letters) in the sequence $\color{cyan}{ABABAB}$ is $\binom{7}{3}$.  
We must place a $C$ between the pair of consecutive $B$'s in the sequences $\color{magenta}{ABABBA}$ and $\color{magenta}{ABBABA}$, which leaves us $\binom{6}{2}$ ways to insert the two remaining $C$'s in the six remaining spaces.
We must place one $C$ between the pair of consecutive $A$'s and another $C$ between the pair of consecutive $B$'s in the sequences $\color{green}{AABABB}$, $\color{green}{AABBAB}$, $\color{green}{ABAABB}$, $\color{green}{ABBAAB}$, leaving five spaces in which to place the remaining $C$.  
Hence, the number of sequences of three $A$'s, three $B$'s, and three $C$'s that begin with $A$ that do not contain consecutive letters that are the same is 
$$1 \cdot 0 + 2 \cdot 1 + 2 \cdot \binom{6}{2} + 4 \cdot \binom{5}{1} + 1 \cdot \binom{7}{3} = 0 + 2 + 30 + 20 + 35 = 87$$
By symmetry, there are also $87$ such sequences that begin with a $B$.  Hence, the number of sequences of three $A$'s, three $B$'s, and three $C$'s that do not contain consecutive letters that are the same is $2 \cdot 87 = 174$.
A: Let $A_i$ be the set of arrangements with at least 2 consecutive letters of type $i$, where $1\le i\le3$.
Then $\displaystyle|\overline{A_1}\cap\overline{A_2}\cap\overline{A_3}|=|S|-\sum_{i}|A_i|+\sum_{i<j}|A_i\cap A_j|-|A_1\cap A_2\cap A_3|,\;\;$
where $\displaystyle|S|=\frac{9!}{3!3!3!}=\color{blue}{1680}$.
1) To find $|A_i|$, we will count the number of arrangements of AA, A, B, B, B, C, C, C and then 
$\;\;\;$adjust for overcounting, which gives $\displaystyle \frac{8!}{3!3!}-\frac{7!}{3!3!}=\color{blue}{980}$.
2) To find $|A_i\cap A_j|$, we will count the number of arrangements of AA, A, BB, B, C, C, C and then
$\;\;\;$adjust for overcounting, which gives $\displaystyle\frac{7!}{3!}-2\cdot\frac{6!}{3!}+\frac{5!}{3!}=\color{blue}{620}$.
3) To find $|A_1\cap A_2\cap A_3|$, we will count the number of arrangements of AA, A, BB, B, CC, C and then
$\;\;\;$adjust for overcounting, which gives $\displaystyle6!-3\cdot5!+3\cdot4!-3!=\color{blue}{426}$.
Therefore $\displaystyle|\overline{A_1}\cap\overline{A_2}\cap\overline{A_3}|=1680-3\cdot980+3\cdot620-426=\color{red}{174}$.
A: A way to find $|a|$.
Let $i,k$ be nonnegative integers and let $j$ be a positive integer (also see the comment of Joriki).
Only looking at $B$'s and $C$'s there are $\binom{6}{3}$ arrangements
for them.
Concerning $iAAAk$: for $i+k=6$ there are $\binom{7}{1}$ solutions. 
Concerning $iAAjAk$: for $i+j+k=6$ there are $\binom{7}{2}$ solutions.
Concerning $iAjAAk$: for $i+j+k=6$ there are $\binom{7}{2}$ solutions.
This lead to $\left|a\right|=\binom{6}{3}\left[\binom{7}{1}+\binom{7}{2}+\binom{7}{2}\right]$.
Above integer $i$ stands for the number of non $A$'s that precede the first $A$ and $j$ and $k$ are sortlike. The context should be enough.
