## Precalculus (6th Edition)

$15$
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. --- $m_{1}=5 \qquad$ ... first name $m_{2}=3 \qquad$ ... middle name Total = $5\times 3$ = $15$