Set theory problem with the inclusion-exclusion principle I started learning set theory, but as for the begin firstly I'm having problems with some notations.
From the lectures we learned about this principle which is as following:
1.$$ |A \cup B|= |A|+|B|$$
The problem with this one is that I don't know where to use it?
EDIT: This part till here has been understood!
Second, from the inclusion and exclusion principle I'm trying to solve these two examples which are: 


*

*Given:
$$
 |A \cup B \cup C|=45 \\
 |A \cap C|=5 \\
 |A \cap B|=4 \\
|B \cap C|=6 \\
 |A \setminus (B \cup C)|=10 \\
 |B \setminus (A \cup C)|=12 \\
 |C \setminus (B \cup A)|=12 $$


Find $|A \cap B \cap C|=?$


*Given:
$$|A \cap B \cap C|=3 \\
  |C|=12 \\ 
|A \setminus (B \cup C)|=10 \\ 
|B \setminus C|=17 \\
|A|=21 \\ 
|U|=49 \\ 
|B \cap C|=5 \\ 
|A \cap C|=5 \\
$$


a) Find:
$$|C \setminus (A \cup B)|=?$$
b) Find:
$$|A \cup (B \cup C)|=?$$
Where $|U| $is the universal set.
I tried using Venn Diagram but it isn't bringing me far, just the first step of organizing the sets.
  P.S I have other examples that I must solve but I think knowing how these work may help me with the others. 
 A: We  partition the  set  $A\cup B\cup C$ into $7$ pairwise disjoint sets
\begin{align*}
A\cup B\cup C&=\left(A\setminus (B\cup C)\right) \cup \left(B\setminus (C\cup A)\right)  \cup \left(C\setminus (A\cup B)\right) \\
&\qquad\cup \left((A\cap B)\setminus C\right) \cup \left((B\cap C)\setminus A\right) \cup \left((C\cap A)\setminus B\right) \\
&\qquad\cup (A\cap B\cap C)
\end{align*}
                      
The strategy  is to  derive  relations consisting solely of one or more of these $7$ atomic subsets and obtain this way the results.
We have
\begin{align*}
|A\cup B\cup C|&=|A\setminus (B\cup C)| + |B\setminus (C\cup A)|  + |C\setminus (A\cup B)| \\
&\qquad+ |(A\cap B)\setminus C| + |(B\cap C)\setminus A| + |(C\cap A)\setminus B|\tag{1} \\
&\qquad+ |A\cap B\cap C|
\end{align*}

First problem:
Since
  \begin{align*}
|A \cup B \cup C|&=45\\
|A \setminus (B \cup C)|&=10\\
|B \setminus (A \cup C)|&=12\\
|C \setminus (B \cup A)|&=12
\end{align*}
   we obtain from (1)
  \begin{align*}
45&=10+12+12\\
&+ |(A\cap B)\setminus C| + |(B\cap C)\setminus A| + |(C\cap A)\setminus B| \\
&\qquad+ |(A\cap B\cap C)|\\
11&=|(A\cap B)\setminus C| + |(B\cap C)\setminus A| + |(C\cap A)\setminus B| \\
&\qquad+ |A\cap B\cap C|\tag{2}
\end{align*}
Since $|A\cap B|=4, |A\cap C|=5$ and  $|B\cap C|=6$ we obtain
\begin{align*}
|A\cap B|&= |(A\cap B)\setminus C| + |A\cap B\cap C|=4\\
|A\cap C|&= |(C\cap A)\setminus B| + |A\cap B\cap C|=5\tag{3}\\
|B\cap C|&= |(B\cap C)\setminus A| + |A\cap B\cap C|=6
\end{align*}
Combining (2)  and (3) we obtain
  \begin{align*}
11&=(4-|A\cap B\cap C|)+(5-|A\cap B\cap C|)+(6-|A\cap B\cap C|)+|A\cap B\cap C|
\end{align*}
We finally conclude
  \begin{align*}
\color{blue}{|A\cap B\cap C|=2}
\end{align*}

A: The Inclusion-Exclusion Principle for two sets is
$$
|A\cup B|=|A|+|B|-|A\cap B|
$$
and for three sets is
$$
|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|
$$

We are given
$$
 |A \cup B \cup C|=45 \\
 |A \cap C|=5 \\
 |A \cap B|=4 \\
|B \cap C|=6\\
 |A \setminus (B \cup C)|=10 \\
 |B \setminus (A \cup C)|=12 \\
 |C \setminus (B \cup A)|=12
$$
For any $A,B,C$, we have $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$ and therefore, Inclusion-Exclusion says that
$$
|A\cap(B\cup C)|=|A\cap B|+|A\cap C|-|A\cap B\cap C|
$$
Also, because $A\cap(B\cup C)$ is disjoint from $A\setminus(B\cup C)$,
$$
|A\cap(B\cup C)|=|A|-|A\setminus(B\cup C)|
$$
Combining the last two equations gives
$$
|A|+|A\cap B\cap C|=|A\cap B|+|A\cap C|+|A\setminus(B\cup C)|
$$
Thus, in particular, reading the blue terms from those given,
$$
\begin{align}
|A|+|A\cap B\cap C|
&=\color{#00F}{|A\cap B|+|A\cap C|+|A\setminus(B\cup C)|}=19\\
|B|+|A\cap B\cap C|
&=\color{#00F}{|B\cap C|+|B\cap A|+|B\setminus(C\cup A)|}=22\\
|C|+|A\cap B\cap C|
&=\color{#00F}{|C\cap A|+|C\cap B|+|C\setminus(A\cup B)|}=23
\end{align}
$$
Summing the last three equations gives
$$
|A|+|B|+|C|+3|A\cap B\cap C|=64
$$
Inclusion-Exclusion gives
$$
\begin{align}
45
&=|A\cup B\cup C|\\
&=\color{#C00}{|A|+|B|+|C|}\color{#090}{-|A\cap B|-|A\cap C|-|B\cap C|}\color{#C00}{+|A\cap B\cap C|}\\
&=\color{#C00}{64}\color{#090}{-15}\color{#C00}{-2|A\cap B\cap C|}
\end{align}
$$
Algebra then gives
$$
|A\cap B\cap C|=2
$$
