Com conhecimento do $Principio \ da \ Inclusão-Exclusão \ ( cardinalidade)$: \begin{matrix} n(A \cup B) & = & n(A) & + & n(B) & - & n(A \cap B) \end{matrix} \begin{matrix} 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) \end{matrix} Assim, podemos escrever: \begin{matrix} n(A) & + & n(B) & - & n(A \cap B) & = & 8 \\ n(A) & + & n(C) & - & n(A \cap C) & = & 9 \\ n(B) & + & n(C) & - & n(B \cap C) & = & 10 \end{matrix} Somando, \begin{matrix} 2n(A) & + & 2n(B) & + & 2n(C) & - & [n(A \cap B) + n(A \cap C) + n(B \cap C) ] & = & 27 \end{matrix} \begin{matrix} \fbox{$\begin{matrix} n(A) & + & n(B) & + & n(C) & - & [n(A \cap B) + n(A \cap C) + n(B \cap C) ] & = & 27 & - & [n(A) + n(B) + n(C)] \end{matrix}$} \end{matrix} Por outro lado, \begin{matrix} n(A) & + & n(B) & + & n(C) & - & [n(A \cap B) + n(A \cap C) + n(B \cap C) ] & + & n(A \cap B \cap C) & = & 11 \\ n(A) & + & n(B) & + & n(C) & - & [n(A \cap B) + n(A \cap C) + n(B \cap C) ] & + & 2 & = & 11 \\ \end{matrix} \begin{matrix} 27 & - & [n(A) + n(B) + n(C)] & + & 2 & = & 11 \\ \end{matrix} \begin{matrix} \fbox{ $ \begin{matrix} n(A) & + & n(B) &+ & n(C) & = & 18 \end{matrix} $ } \end{matrix} \begin{matrix} Letra \ (D) \end{matrix}