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Returning to our example above, we have \(\card{A} = 12\text{,}\) \(\card{B} = 5\text{,}\) \(\card{C} = 8\text{.}\) We also have \(\card{A \cap B} = 2\text{,}\) \(\card{A \cap C} = 6\text{,}\) \(\card{B \cap C} = 3\text{,}\) and \(\card{A \cap B \cap C} = 1\text{.}\) Therefore:

\begin{equation*} \card{A \cup B \cup C} = 12 + 5 + 8 - 2 - 6 - 3 + 1 = 15\text{.} \end{equation*}
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