## Precalculus (6th Edition)

$303,600$.
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. --- $m_{1}=25$ ... any of the 25 boys can receive the first type of award $m_{2}=24$ ... of the 24 remaining, one will receive the second award $m_{3}=23$ ... of the 23 remaining, one will receive the third award, $m_{4}=22$ ... 22 remain to choose from for the 4th award. Total = $25\times 24\times 23\times 22$ = $303,600.$ Another approach: Permutations, taking r=4 from n=25 boys. P(25,4)= $303,600$.