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A universal set [tex]\( U \)[/tex] consists of 23 elements. If sets [tex]\( A, B \)[/tex], and [tex]\( C \)[/tex] are proper subsets of [tex]\( U \)[/tex] and [tex]\( n(U) = 23 \)[/tex], [tex]\( n(A \cap B) = n(A \cap C) = n(B \cap C) = 9 \)[/tex], [tex]\( n(A \cap B \cap C) = 4 \)[/tex], and [tex]\( n(A \cup B \cup C) = 19 \)[/tex], determine each of the following.

a) [tex]\( n(A \cup B) = \)[/tex] [tex]\(\square\)[/tex]
(Simplify your answer.)

b) [tex]\( n\left(A^{\prime} \cup C\right) = \)[/tex] [tex]\(\square\)[/tex]
(Simplify your answer.)

c) [tex]\( n(A \cap B)^{\prime} = \)[/tex] [tex]\(\square\)[/tex]
(Simplify your answer.)

Sagot :

GinnyAnswer
To solve the given problem, we need to use principles from set theory and properties such as union, intersection, and complement of sets. Let's go step by step.

Given:
- Universal set [tex]\( U \)[/tex] has 23 elements: [tex]\( n(U) = 23 \)[/tex]
- [tex]\( n(A \cap B) = 9 \)[/tex]
- [tex]\( n(A \cap C) = 9 \)[/tex]
- [tex]\( n(B \cap C) = 9 \)[/tex]
- [tex]\( n(A \cap B \cap C) = 4 \)[/tex]
- [tex]\( n(A \cup B \cup C) = 19 \)[/tex]

### Part (a) [tex]\( n(A \cup B) \)[/tex]

Since [tex]\( n(A \cup B \cup C) = 19 \)[/tex], this is already provided and pertains to the union of sets [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex].

To find [tex]\( n(A \cup B) \)[/tex], we use the formula for the union of two sets:
[tex]\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \][/tex]

However, since the original question specifies [tex]\( A \cup B = 19 \)[/tex] directly:
[tex]\[ n(A \cup B) = 19 \][/tex]

### Part (b) [tex]\( n(A' \cup C) \)[/tex]

The complement [tex]\( A' \)[/tex] (also written as [tex]\( A^c \)[/tex]) means the set of elements in [tex]\( U \)[/tex] that are not in [tex]\( A \)[/tex]. Here's the formula we will use for the union involving a complement:
[tex]\[ n(A' \cup C) = n(U) - n(A \cap C') = n(U) - n(A) + n(A \cap C) - n(C) \][/tex]

Since [tex]\( n(A \cap B \cap C) = 4 \)[/tex], applying the formula: [tex]\( n(A \cap C) \)[/tex] includes elements in their common intersection [tex]\(n(A \cap B \cap C)\)[/tex], so:
[tex]\[ n(A) + n(C) - n(A \cap C) + n(A' \cap C) = n(U) \][/tex]

We know from De Morgan's laws:
[tex]\[ A' \cup C = (A \cap C') \cup C = U - A \][/tex]

Thus:
[tex]\[ n(A' \cup C) = n(U) - n(A) + n(C) - n(A \cap B \cap C)\][/tex]

Since :
[tex]\[ n(U) = 23 - 19=4\][/tex]

### Part (c) [tex]\( n(A \cap B)' \)[/tex]

The complement of [tex]\( n(A \cap B) \)[/tex] is the total number of elements not in [tex]\( A \cap B \)[/tex], given by:
[tex]\[ n(A \cap B)' = n(U) - n(A \cap B) \][/tex]

Given:
[tex]\[ n(U) = 23 \][/tex]
[tex]\[ n(A \cap B) = 9 \][/tex]

So:
[tex]\[ n(A \cap B)' = 23 - 9 = 14 \][/tex]

To summarize:
[tex]\[ \begin{align*} a) \quad &n(A \cup B) = 19 \\ b) \quad &n(A' \cup C) = 4 \\ c) \quad &n(A \cap B)' = 14 \end{align*} \][/tex]
These are the simplified answers to the given questions.
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