1. **Problem Statement:** Find the following from the Venn diagram with sets A, B, C, and universal set 𝓔: i) $n(A \cup B)$ ii) $n(A' \cap B')$ iii) $n((A \cap B) \cup C)$ 2. **Recall formulas and rules:** - $A \cup B$ means all elements in A or B or both. - $A'$ means elements not in A (complement of A). - $A' \cap B'$ means elements not in A and not in B. - $A \cap B$ means elements common to both A and B. - Union and intersection operations follow set algebra rules. 3. **Given numbers in regions:** - Only A: 12 - Only B: 9 - Only C: 4 - A and B only (not C): 5 - A and C only (not B): 10 - B and C only (not A): 3 - A and B and C: 6 - Outside all sets (in 𝓔 only): 8 4. **Calculate $n(A \cup B)$:** $A \cup B$ includes all elements in A or B or both, so sum all regions in A or B: $$n(A \cup B) = 12 + 5 + 10 + 9 + 3 + 6$$ Note: 10 and 3 are in C intersections but also in A or B respectively. But 10 is in A and C only (so in A), 3 is in B and C only (so in B). So sum is: $$12 + 5 + 10 + 9 + 3 + 6 = 45$$ 5. **Calculate $n(A' \cap B')$:** This is elements not in A and not in B, so elements outside A and B. From the diagram, these are elements only in C (4) and outside all sets (8): $$n(A' \cap B') = 4 + 8 = 12$$ 6. **Calculate $n((A \cap B) \cup C)$:** First find $A \cap B$ which includes elements in both A and B: - A and B only (not C): 5 - A and B and C: 6 So, $$n(A \cap B) = 5 + 6 = 11$$ Now union with C (all elements in C): C includes: - Only C: 4 - A and C only: 10 - B and C only: 3 - A and B and C: 6 So, $$n(C) = 4 + 10 + 3 + 6 = 23$$ Union $ (A \cap B) \cup C$ includes all elements in either $A \cap B$ or C: Since $A \cap B$ elements 5 and 6 are already counted in C (6 is in all three, 5 is not in C), we add 5 to C's total: $$n((A \cap B) \cup C) = n(C) + n(A \cap B \text{ only}) = 23 + 5 = 28$$ **Final answers:** - i) $n(A \cup B) = 45$ - ii) $n(A' \cap B') = 12$ - iii) $n((A \cap B) \cup C) = 28$