https://file.aiquickdraw.com/mathbot/file/1726953157176tkknwup9.jpg

Let's analyze and solve the problem step by step based on the given sets.

### 1st Row (Sets \(A\) and \(B\)):

- \(|A| = 4\), \(|B| = 2\), and \(|A \cap B| = 0\).
  
Since \(|A \cap B| = 0\), the sets \(A\) and \(B\) are disjoint, meaning they have no elements in common. 

- \(|A \cup B|\): The union of disjoint sets is simply the sum of their cardinalities:
  \[
  |A \cup B| = |A| + |B| = 4 + 2 = 6
  \]
  
- \(|A \setminus B|\): The difference \(|A \setminus B|\) refers to the number of elements in \(A\) that are not in \(B\). Since the sets are disjoint, all elements of \(A\) are not in \(B\), so:
  \[
  |A \setminus B| = |A| = 4
  \]

### 2nd Row (Sets \(C\) and \(D\)):

- \(|C| = 8\), \(|D| = 4\), and \(|C \cap D| = 3\).

- \(|C \cup D|\): The formula for the union of two sets is:
  \[
  |C \cup D| = |C| + |D| - |C \cap D| = 8 + 4 - 3 = 9
  \]
  
- \(|C \setminus D|\): The difference \(|C \setminus D|\) refers to the number of elements in \(C\) that are not in \(D\). This is:
  \[
  |C \setminus D| = |C| - |C \cap D| = 8 - 3 = 5
  \]

### 3rd Row (Sets \(E\) and \(F\)):

- \(|E| = 13\), \(|F| = 20\), and \(|E \cup F| = 26\).

We are not given \(|E \cap F|\) directly, but we can deduce it from the union formula:

\[
|E \cup F| = |E| + |F| - |E \cap F|
\]

Substituting the values:

\[
26 = 13 + 20 - |E \cap F|
\]

Solving for \(|E \cap F|\):

\[
|E \cap F| = 33 - 26 = 7
\]

- \(|E \setminus F|\): The difference \(|E \setminus F|\) refers to the number of elements in \(E\) that are not in \(F\). This is:
  \[
  |E \setminus F| = |E| - |E \cap F| = 13 - 7 = 6
  \]

### Final Answers:

1. \(|A \cup B| = 6\), \(|A \setminus B| = 4\)
2. \(|C \cup D| = 9\), \(|C \setminus D| = 5\)
3. \(|E \cap F| = 7\), \(|E \setminus F| = 6\)

Would you like any further details or explanations?

Here are 5 related questions:
1. How would the solution change if \( |A \cap B| \neq 0 \)?
2. Can you express the union and intersection formulas using Venn diagrams?
3. What is the meaning of \( |A \setminus B| \) in practical terms?
4. How would you calculate \( |A \cup B| \) if more than two sets are involved?
5. How does set theory apply to probability?

**Tip**: Always use the union formula \( |A \cup B| = |A| + |B| - |A \cap B| \) when sets overlap.

The sets A and B are such that |A| = 4, |B| = 2, and |A ∩ B| = 0. Then |A ∪ B| is ___, and |A \ B| is ___. The sets C and D are such that |C| = 8, |D| = 4, and |C ∩ D| = 3. Then |C ∪ D| is ___, and |C \ D| is ___. The sets E and F are such that |E| = 13, |F| = 20, and |E ∪ F| = 26. Then |E ∩ F| is ___, and |E \ F| is ___.

Solve a set theory problem involving unions, intersections, and set differences. The problem provides cardinalities of sets A, B, C, D, E, and F and requires calculating union and set difference for each pair. This problem explores key concepts in set theory, such as disjoint sets, overlapping sets, and cardinality rules.

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