a) $\#\left( A \right)=6$
b) $\#\left( B \right)=9$
c) $\#\left( A\cap B \right)=3$
$\#\left( A\cup B
\right)=\#\left( A \right)+\#\left( B \right)-\#\left( A\cap B \right)$
Luego en
nuestro caso:
\[\Rightarrow \#\left(
A\cup B \right)=\#\left( A \right)+\underbrace{\#\left( A \right)+\#\left(
A\cap B \right)}_{\#\left( B \right)}-\#\left( A\cap B \right)=12\Rightarrow \]
\[\Rightarrow 2\#\left( A \right)=12\quad
\Rightarrow \quad \#\left( A \right)=\frac{12}{2}\quad \Rightarrow \quad
\#\left( A \right)=6\]
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