1. **Problem statement:** Given sets and their cardinalities, find possible elements for sets B, C, a value for x in the complement of A, and calculate $n(D \cup C)$. 2. **Recall definitions:** - $n(X)$ is the number of elements in set $X$. - $A \cap B$ is the intersection of $A$ and $B$ (elements common to both). - $C \subset B$ means $C$ is a subset of $B$. - $A'$ is the complement of $A$ (elements not in $A$). - $n(D \cup C) = n(D) + n(C) - n(D \cap C)$. 3. **Given:** - $A = \{2,5,9,13,15,16\}$ with $n(A) = 6$. - $n(B) = 4$, $n(A \cap B) = 3$. - $n(C) = 2$, $C \subset B$. - $n(D) = 9$, $n(D \cap C) = 1$. 4. **Find B:** Since $n(B) = 4$ and $n(A \cap B) = 3$, $B$ must have 3 elements from $A$ and 1 element not in $A$. Choose $B = \{2,5,9,20\}$ where $20 \notin A$. 5. **Find C:** $C$ is a subset of $B$ with 2 elements. Choose $C = \{2,20\}$. 6. **Find $x \in A'$:** $x$ is an element not in $A$. Choose $x = 1$ (since $1 \notin A$). 7. **Calculate $n(D \cup C)$:** Given $n(D) = 9$, $n(C) = 2$, and $n(D \cap C) = 1$. Use formula: $$ n(D \cup C) = n(D) + n(C) - n(D \cap C) = 9 + 2 - 1 = 10 $$ **Final answers:** - $B = \{2,5,9,20\}$ - $C = \{2,20\}$ - $x = 1$ - $n(D \cup C) = 10$