I see that you've uploaded an image with a set of mathematical problems in Korean, related to set theory and unions, intersections, and complements. I’ll translate and solve the problems one by one for you. Let me analyze and provide solutions.

Problem 1

두 집합 $A$, $B$에 대하여
$n(A) = 18$, $n(B) = 15$, $n(A \cup B) = 26$일 때, $n(A \cap B)$를 구하시오.

Solution: We use the formula for the union of two sets:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$ Substitute the given values:
$26 = 18 + 15 - n(A \cap B)$ $26 = 33 - n(A \cap B)$ $n(A \cap B) = 33 - 26 = 7$

Answer: $n(A \cap B) = 7$

Problem 2

전체집합 $U$의 부분집합 $A$에 대하여
$n(A^c) = 17$, $n(A) = 7$일 때, $n(U)$의 값을 구하시오.

Solution: The complement rule tells us that:
$n(U) = n(A) + n(A^c)$ Substitute the given values:
$n(U) = 7 + 17 = 24$

Answer: $n(U) = 24$

Problem 3

두 집합 $A$, $B$에 대하여
$n(A) = 15$, $n(A \cap B) = 6$일 때, $n(A - B)$의 값을 구하시오.

Solution: The difference of sets $A - B$ is defined as the elements in $A$ but not in $B$:
$n(A - B) = n(A) - n(A \cap B)$ Substitute the given values:
$n(A - B) = 15 - 6 = 9$

Answer: $n(A - B) = 9$

Problem 4

전체집합 $U$의 두 부분집합 $A, B$에 대하여
$n(U) = 35$, $n(A) = 16$, $n(B) = 19$, $n(A \cap B) = 6$일 때, $n(B - A)$의 값을 구하시오.

Solution: We calculate $n(B - A)$, which is the number of elements in $B$ but not in $A$:
$n(B - A) = n(B) - n(A \cap B)$ Substitute the given values:
$n(B - A) = 19 - 6 = 13$

Answer: $n(B - A) = 13$

Let me know if you'd like further explanation or assistance!

Follow-up Questions:

1. Can you explain the complement rule in set theory in more detail?
2. How can we visualize the relationship between $A$, $B$, and $A \cup B$ with a Venn diagram?
3. What are the key differences between $A - B$ and $A \cap B$?
4. How would the solution to Problem 4 change if $n(U)$ wasn't given?
5. What is the importance of understanding set theory in real-world applications?

Tip:

When solving problems involving set theory, always start by writing down the relevant formula and substitute values carefully to avoid errors.