Use inclusion/exclusion to find the number of derangements of the string aabbcc There is a problem exactly like the one I asked, however I am still stuck and still need help. So I tried to edit the question with the work you see below before deciding to ask this question again, so people can actually help me. So here is my attempt. 

Let A denotes where aa occurs, B denotes where bb occurs, C denotes where cc occurs, and |U|=$\frac{6!}{2^3}$. Then $|A\cup B\cup C|=|A|+|B|+|C|-(|AB|+|AC|+|BC|)+|ABC|$.

For |A|+|B|+|C| glue aa together for |A|, bb for |B| and cc for |C| 

|A|: $\frac{5!}{2^2}$ |B|: $\frac{5!}{2^2}$  |C|: $\frac{5!}{2^2}$ $\implies$ |A|+|B|+|C|=$3\left(\frac{5!}{2^2}\right)$

For |AB|+|AC|+|BC|=$3\left(\frac{4!}{2}\right)$ 
For |ABC|= 3!

So |U|-$|A\cup B\cup C| = \frac{6!}{2^3}-\left[\frac{5!}{2^2}-3\left(\frac{4!}{2}\right)+3!\right] = 30$

Now thanks to the people who already answered this we know that the actual answer is 10. 
As you all can see I am doing something wrong, but I not sure what. By using the Inclusion-Exclusion Principle please help me figure out how I can get the correct answer.
 A: In a derangement, no letter is left in its original position.  You only considered the possibility that no pair of letters was left in its original position.
There are 
$$\frac{6!}{2!2!2!}$$
distinguishable arrangements of the letters of the word $aabbcc$.
In a derangement, no letter is left in its original position.  Therefore, we must exclude those cases in which one or more letters is left in its original position.
Define $A_1$ to be the set of arrangements in which the first $a$ is left in its original position.  Define $A_2$ to be the set of arrangements in which the second $a$ is left in its original position.  Define $B_1$, $B_2$, $C_1$, and $C_2$ analogously for the letters $b$ and $c$.
one letter is left in its original position:  Consider $|A_1|$.  Since the first $a$ is in its original position, we are left with an arrangement of the set $\{a, b, b, c, c\}$, so 
$$|A_1| = \frac{5!}{2!2!}$$
By symmetry, 
$$|A_1| = |A_2| = |B_1| = |B_2| = |C_1| = |C_2|$$
Hence, there are a total of 
$$\binom{6}{1}\frac{5!}{2!2!}$$
arrangements in which one letter is in its original position.
two letters are in their original positions:  We have two cases to consider.  The letters are the same, or they are different.
two identical letters are in their original positions:  Consider $|A_1 \cap A_2|$.  If both $a$s are in their original positions, we are left with an arrangement of the set $\{b, b, c, c\}$.  Hence,
$$|A_1 \cap A_2| = \frac{4!}{2!2!}$$
By symmetry, 
$$|A_1 \cap A_2| = |B_1 \cap B_2| = |C_1 \cap C_2|$$
Hence, there are 
$$\binom{3}{1}\frac{4!}{2!2!}$$
arrangements in which two identical letters are in their original positions.
two different letters are in their original positions:  Consider $|A_1 \cap B_1|$.  We are left with an arrangement of the set $\{a, b, c, c\}$.  Hence, 
$$|A_1 \cap B_1| = \frac{4!}{2!}$$
There are $\binom{3}{2}$ ways to pick two different letters to be fixed points and two ways to choose the position of the fixed point for each of those letters.  Hence, there are 
$$\binom{3}{2}\binom{2}{1}^2\frac{4!}{2!}$$
arrangement with two different letters in their original positions.
three letters are in their original positions:  We have two cases to consider.  Either a pair of identical letters and one other letter are in their original positions, or three different letters are in their original positions.  
a pair of identical letters and one other letter are in their original positions:  Consider $|A_1 \cap A_2 \cap B_1|$.  Since both $a$s and the first $b$ are fixed, we have an arrangement of the set $\{b, c, c\}$.  Hence,
$$|A_1 \cap A_2 \cap B_1| = \frac{3!}{2!}$$
By symmetry, there are three ways of picking the repeated letter, two ways of choosing the letter for the other fixed point, and two ways of choosing the position of the fixed point for that letter.  Hence, there are 
$$\binom{3}{1}\binom{2}{1}\binom{2}{1}\frac{3!}{2!}$$
arrangements in which there are three letters in their original positions, in which two of the fixed points are identical letters.
three different letters are in their original positions:  Consider $|A_1 \cap B_1 \cap C_1|$.  Since the first $a$, first $b$, and first $c$ are fixed points, we are left with an arrangement of $\{a, b, c\}$.  Hence, 
$$|A_1 \cap B_1 \cap C_1| = 3!$$ 
By symmetry, there are two ways to pick the fixed point for each of the three letters.  Hence, there are 
$$\binom{2}{1}^33!$$
arrangements in which three different letters are fixed points.
four letters are in their original positions:  Again, there are two cases.  Either two pairs of identical letters are fixed points, or one pair of identical letters and two different letters are fixed points. 
two pairs of identical letters are in their original positions:  Consider 
$|A_1 \cap A_2 \cap B_1 \cap B_2|$. Since both $a$s and both $b$s are in their original positions, we are left with an arrangement of $\{c, c\}$.  Hence, 
$$|A_1 \cap A_2 \cap B_1 \cap B_2| = \frac{2!}{2!} = 1$$
By symmetry, there are $\binom{3}{2}$ ways of picking two pairs of identical letters to be fixed points.  Hence, there are 
$$\binom{3}{2}$$
arrangements in which two pairs of identical letters are fixed points.
one pair of identical letters and two different letters are fixed points:  Consider $|A_1 \cap A_2 \cap B_1 \cap C_1|$.  Since both $a$s, the first $b$, and first $c$ are fixed points, we are left with an arrangement of $\{b, c\}$.  Hence, 
$$|A_1 \cap A_2 \cap B_1 \cap C_1| = 2!$$
Since there are three ways of choosing the repeated letter and two ways of choosing the position of the fixed points for the other two letters, there are 
$$\binom{3}{1}\binom{2}{1}^22!$$
arrangements in which a pair of identical letters and two different letters are fixed points.
five letters are in their original positions:  Consider $|A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_1|$.  Since both $a$s, both $b$s, and the first $c$ are in their original positions, we have an arrangement of $\{c\}$.  Hence, 
$$|A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_1| = 1$$
By symmetry, there are $\binom{6}{5}$ ways to select the fixed point.  Hence, there are 
$$\binom{6}{5}$$
arrangements with five fixed points.
six letters are in their original positions:  Since all the letters are in their original positions, 
$$|A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_1 \cap C_2| = 1$$
Thus, by the Inclusion-Exclusion Principle, the number of derangements of $aabbcc$ is 
$$\frac{6!}{2!2!2!} - \binom{6}{1}\frac{5!}{2!2!} + \binom{3}{1}\frac{4!}{2!2!} + \binom{3}{2}\binom{2}{1}^2\frac{4!}{2!} - \binom{3}{1}\binom{2}{1}\binom{2}{1}\frac{3!}{2!} - \binom{2}{1}^33! + \binom{3}{2} + \binom{3}{1}\binom{2}{1}^22! - \binom{6}{5} + \binom{6}{6}$$ 
A: Since there are no a's in places 1 and 2, there are three possibilities for these places:
$\textbf{1)}$ If there are 2 b's in places 1-2, then the c's are in places 3-4, so we have bbccaa.
$\textbf{2)}$ If there are 2 c's in places 1-2, then the b's are in places 5-6, so we have ccaabb.
$\textbf{3)}$ If there is 1 b and 1 c in places 1-2, then there are 2 ways to order the b and c, 2 choices for placing the remaining c (place 3 or place 4), and 2 choices for placing the remaining b (place 5 or place 6).
This gives a total of $1+1+2\cdot2\cdot2=\color{red}{10}$ possibilities.
