1. **Problem statement:** Given a universal set $E$ and three sets $A$, $B$, and $C$ with elements distributed as follows: - $6$ elements only in $A$ - $3$ elements in $A \cap B$ - $8$ elements only in $B$ - $2$ elements in $B \cap C$ - $5$ elements only in $C$ - $4$ elements outside $A$, $B$, and $C$ Find: (a) $n(A \cup B)$ (b) $n(A \cap B)$ (c) $n(B \cap C')$ (d) $n(A' \cup B' \cup C')$ 2. **Formulas and rules:** - Union: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ - Intersection: $n(A \cap B)$ is given or can be found by overlapping regions. - Complement: $C'$ means elements not in $C$. - De Morgan's law: $A' \cup B' \cup C' = (A \cap B \cap C)'$ 3. **Calculate each part:** (a) $n(A \cup B)$: - $n(A) = 6 + 3 = 9$ (elements only in $A$ plus $A \cap B$) - $n(B) = 8 + 3 + 2 = 13$ (elements only in $B$, $A \cap B$, and $B \cap C$) - $n(A \cap B) = 3$ (given) So, $$ n(A \cup B) = 9 + 13 - 3 = 19 $$ (b) $n(A \cap B)$: - Given as $3$ (c) $n(B \cap C')$: - $C'$ means not in $C$ - Elements in $B$ but not in $C$ are those in $B$ excluding $B \cap C$ - So, $n(B \cap C') = 8 + 3 = 11$ (d) $n(A' \cup B' \cup C')$: - By De Morgan's law, this is the complement of $A \cap B \cap C$ - Since there is no region where all three sets overlap, $n(A \cap B \cap C) = 0$ - Total elements in $E$ are $6 + 3 + 8 + 2 + 5 + 4 = 28$ Therefore, $$ n(A' \cup B' \cup C') = 28 - 0 = 28 $$ **Final answers:** - (a) $19$ - (b) $3$ - (c) $11$ - (d) $28$