Let [tex]\( U \)[/tex] be the universal set, where [tex]\( U = \{1,2,3, \ldots, 24, 25\} \)[/tex].

Let [tex]\( A \)[/tex] and [tex]\( B \)[/tex] be subsets of [tex]\( U \)[/tex], where

[tex]\[
A = \{4, 6, 16, 17, 19, 20, 22, 24\} \\
B = \{1, 2, 3, 4, 8, 9, 10, 11, 14, 23, 24, 25\}
\][/tex]

List the elements in the set [tex]\( A \cap B \)[/tex]:
[tex]\[ A \cap B = \{4, 24\} \][/tex]

List the elements in the set [tex]\( A \cup B \)[/tex]:
[tex]\[ A \cup B = \{1, 2, 3, 4, 6, 8, 9, 10, 11, 14, 16, 17, 19, 20, 22, 23, 24, 25\} \][/tex]

List the elements in the set [tex]\( (A \cup B)^C \)[/tex]:
[tex]\[ (A \cup B)^C = \{5, 7, 12, 13, 15, 18, 21\} \][/tex]

List the elements in the set [tex]\( A^C \cap B^C \)[/tex]:
[tex]\[ A^C \cap B^C = \{5, 7, 12, 13, 15, 18, 21\} \][/tex]



Answer :

Let's solve each part of this problem step by step.

First, let's recall the universal set [tex]\( U \)[/tex] and the subsets [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ U = \{1, 2, 3, \ldots, 24, 25\} \][/tex]
[tex]\[ A = \{4, 6, 16, 17, 19, 20, 22, 24\} \][/tex]
[tex]\[ B = \{1, 2, 3, 4, 8, 9, 10, 11, 14, 23, 24, 25\} \][/tex]

### 1. Finding [tex]\( A \cap B \)[/tex]:
The intersection of sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is the set of elements that are in both [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
[tex]\[ A \cap B = \{ \text{elements in both } A \text{ and } B \} \][/tex]
Looking at sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we see that the common elements are [tex]\( 4 \)[/tex] and [tex]\( 24 \)[/tex].
Thus,
[tex]\[ A \cap B = \{4, 24\} \][/tex]

### 2. Finding [tex]\( A \cup B \)[/tex]:
The union of sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is the set of elements that are in [tex]\( A \)[/tex] or [tex]\( B \)[/tex] or in both.
[tex]\[ A \cup B = \{ \text{all unique elements in either } A \text{ or } B \text{ or both} \} \][/tex]
Combining all unique elements from [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we have:
[tex]\[ A \cup B = \{1, 2, 3, 4, 6, 8, 9, 10, 11, 14, 16, 17, 19, 20, 22, 23, 24, 25\} \][/tex]

### 3. Finding [tex]\( (A \cup B)^C \)[/tex]:
The complement of the union [tex]\( (A \cup B)^C \)[/tex] is the set of elements in the universal set [tex]\( U \)[/tex] that are not in [tex]\( A \cup B \)[/tex].
[tex]\[ (A \cup B)^C = U - (A \cup B) \][/tex]
We list all elements from [tex]\( 1 \)[/tex] to [tex]\( 25 \)[/tex] and subtract the elements in [tex]\( A \cup B \)[/tex].
The elements in [tex]\( U \)[/tex] that are not in [tex]\( A \cup B \)[/tex] are:
[tex]\[ (A \cup B)^C = \{5, 7, 12, 13, 15, 18, 21\} \][/tex]

### 4. Finding [tex]\( A^C \cap B^C \)[/tex]:
To find [tex]\( A^C \cap B^C \)[/tex], we first find [tex]\( A^C \)[/tex] and [tex]\( B^C \)[/tex].

[tex]\( A^C \)[/tex] is the set of elements in [tex]\( U \)[/tex] not in [tex]\( A \)[/tex].
[tex]\[ A^C = U - A = \{1, 2, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 18, 21, 23, 25\} \][/tex]

[tex]\( B^C \)[/tex] is the set of elements in [tex]\( U \)[/tex] not in [tex]\( B \)[/tex].
[tex]\[ B^C = U - B = \{5, 6, 7, 12, 13, 15, 16, 17, 18, 19, 20, 21, 22\} \][/tex]

The intersection [tex]\( A^C \cap B^C \)[/tex] is the set of elements that are in both [tex]\( A^C \)[/tex] and [tex]\( B^C \)[/tex]:
[tex]\[ A^C \cap B^C = \{5, 7, 12, 13, 15, 18, 21\} \][/tex]

### Summary:
- [tex]\( A \cap B = \{4, 24\} \)[/tex]
- [tex]\( A \cup B = \{1, 2, 3, 4, 6, 8, 9, 10, 11, 14, 16, 17, 19, 20, 22, 23, 24, 25\} \)[/tex]
- [tex]\( (A \cup B)^C = \{5, 7, 12, 13, 15, 18, 21\} \)[/tex]
- [tex]\( A^C \cap B^C = \{5, 7, 12, 13, 15, 18, 21\} \)[/tex]

These are the elements in each of the respective sets as required.