Precalculus (6th Edition)

$12$ ways
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. --- 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