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Step-by-Step Solution
Step 1: Understand the Sets B and C
We have a set A = \{1, 2, 3, 4, 5, 6, 7\} . Two subsets of the power set of A are defined as follows:
B = \{\,T \subseteq A : \text{either } 1 \notin T \text{ or } 2 \in T\,\}
In other words, any subset that does not contain 1, or if it does contain 1, it must also contain 2.
C = \{\,T \subseteq A : \text{the sum of the elements of } T \text{ is prime}\} .
Step 2: Use De Morgan's Law to Express (B ∪ C)
We want n(B \cup C) , the number of elements in B \cup C . By De Morgan’s laws for sets:
(B \cup C)' = B' \cap C'.
Therefore:
n(B \cup C) = n(\Omega) - n\bigl(B' \cap C'\bigr),
where \Omega is the entire power set of A , and n(\Omega) = 2^7 = 128.
Step 3: Interpret B' and C'
B' : This is the set of all subsets of A not in B . So B' consists of
subsets that must contain 1 and must not contain 2.
Symbolically, B' = \{\, T \subseteq A : 1 \in T, \, 2 \notin T \}\,.
C' : This is the set of all subsets of A whose sum of elements is not prime.
Symbolically, C' = \{\, T \subseteq A : \sum(\text{elements of } T)\text{ is not prime} \}\,.
Hence, B' \cap C' consists of all subsets that contain 1, do not contain 2, and have a (non-prime) sum of elements.
Step 4: Reduce the Counting to Subsets of {3,4,5,6,7}
Since any subset in B' already has 1 and does not have 2, the only “free” elements left to choose from are
\{3,4,5,6,7\} . Let S be a subset of \{3,4,5,6,7\} . Then the total subset of A is
\{1\} \cup S . The sum of this subset is (1 + \text{sum}(S)) .
Thus counting B' \cap C' translates to counting all subsets S \subseteq \{3,4,5,6,7\} for which 1 + \text{sum}(S) is composite.
Step 5: Count the Number of Such Subsets
According to the provided enumeration (and one can verify each possible subset sum), the total number of subsets
S \subseteq \{3,4,5,6,7\} for which 1 + \text{sum}(S) is composite is 21. Here is a brief reasoning logic
(though enumerations in detail can be done systematically):
Consider all possible subset sizes (from 0 elements to 5 elements).
Check which sums lead to 1 + \text{sum}(S) being composite.
Count those valid subsets in each category and sum the result to get 21.
Step 6: Find n(B ∪ C)
We now compute:
n(B' \cap C') = 21 \quad \Longrightarrow \quad n(B \cup C) = n(\Omega) - n(B' \cap C') = 128 - 21 = 107.
Hence, the number of elements in the set B \cup C is \boxed{107} .
Final Answer
The required number of elements in B \cup C is 107.
