Set theory problem with the inclusion-exclusion principle

Question

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:

1. 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 $|A \cap B \cap C|=?$

2. 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.

Answer 1

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*}

Answer 2

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

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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 $$ |A\cap B\cap C|=2 $$