We are given:

• Universal set $S = \{1, 2, 3, ..., 20\}$
• Set $A = \{1, 3, 4, 8, 9, 12, 13, 15, 16, 17, 19, 20\}$
• Set $B = \{6, 9, 10, 11, 12, 13, 14, 15, 16, 17, 20\}$

Now, we need to find the following:

1. Number of elements in the set $A \cup B$ (the union of A and B): $n(A \cup B)$
2. Number of elements in the set $A \cap B$ (the intersection of A and B): $n(A \cap B)$

Step 1: Find the union $A \cup B$

The union of two sets consists of all unique elements from both sets. We simply combine all the elements in $A$ and $B$ without repetition.

• $A = \{1, 3, 4, 8, 9, 12, 13, 15, 16, 17, 19, 20\}$
• $B = \{6, 9, 10, 11, 12, 13, 14, 15, 16, 17, 20\}$

Taking the union:

$A \cup B = \{1, 3, 4, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20\}$

The number of elements in $A \cup B$ is the count of unique elements in the union:

$n(A \cup B) = 16$

Step 2: Find the intersection $A \cap B$

The intersection of two sets consists of only the elements that are present in both sets.

• $A = \{1, 3, 4, 8, 9, 12, 13, 15, 16, 17, 19, 20\}$
• $B = \{6, 9, 10, 11, 12, 13, 14, 15, 16, 17, 20\}$

Taking the intersection:

$A \cap B = \{9, 12, 13, 15, 16, 17, 20\}$

The number of elements in $A \cap B$ is the count of elements in the intersection:

$n(A \cap B) = 7$

Final Answers:

• $n(A \cup B) = 16$
• $n(A \cap B) = 7$

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Would you like further explanations or have additional questions on this topic? Here are some related questions you might find helpful:

1. What is the difference between the union and the intersection of sets?
2. How can you use a Venn diagram to represent the union and intersection of sets?
3. What happens if the sets $A$ and $B$ have no elements in common (i.e., they are disjoint)?
4. How do we find the complement of a set, and what is the complement of $A \cup B$ in this case?
5. Can the number of elements in the intersection $A \cap B$ be greater than the number of elements in $A \cup B$?

Tip: To avoid counting duplicate elements when finding the union of sets, always list the elements from both sets and eliminate repetitions.