Answer
$12$ ways
Work Step by Step
Apply the Fundamental Principle of Counting
If $n$ independent events occur, with $m_{1}$ ways for event 1 to occur,
$m_{2}$ ways for event 2 to occur,
$\ldots$ and $m_{n}$ ways for event $n$ to occur,
then there are $m_{1}\cdot m_{2}\cdot\cdots\cdot m_{n}$ different ways for all $n$ events to occur.
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There are $m_{1}=6$ ways to obtain an outcome on the die.
There are $m_{2}=2$ ways to obtain an outcome on the coin.
Total = $6\times 2=12$ ways