1. **Problem statement:** Given sets A, B, and C with the following cardinalities: $n(A) = 18$, $n(B) = 21$, $n(C) = 22$, $n(A \cap B) = 9$, $n(A \cap C) = 7$, $n(B \cap C) = 11$, and $n(A \cap B \cap C) = 2$. Find: a) $n(B \cup C)$ b) $n(A \cup B)$ c) $n(A \cup C)$ d) $n(A \cup B \cup C)$ 2. **Formula used:** For any two sets $X$ and $Y$, $$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$$ For three sets $A$, $B$, and $C$, $$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)$$ 3. **Calculations:** a) Calculate $n(B \cup C)$: $$n(B \cup C) = n(B) + n(C) - n(B \cap C) = 21 + 22 - 11 = 32$$ b) Calculate $n(A \cup B)$: $$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 18 + 21 - 9 = 30$$ c) Calculate $n(A \cup C)$: $$n(A \cup C) = n(A) + n(C) - n(A \cap C) = 18 + 22 - 7 = 33$$ d) Calculate $n(A \cup B \cup C)$: $$n(A \cup B \cup C) = 18 + 21 + 22 - 9 - 7 - 11 + 2 = 36$$ 4. **Final answers:** a) $n(B \cup C) = 32$ b) $n(A \cup B) = 30$ c) $n(A \cup C) = 33$ d) $n(A \cup B \cup C) = 36$