Answer
There are $303,600$ possible outcomes.
Work Step by Step
The order in which the balls are placed matters and no ball can be selected twice.
Permutation of 25 distinct balls taken 4 at a time:
$_{25}P_{4}=\frac{25!}{(25-4)!}=\frac{25!}{21!}$
But, $25!=25\times24\times23\times22\times(21\times20\times19\times ... \times3\times2\times1)=25\times24\times23\times22\times21!$
$_{25}P_{4}=\frac{25\times24\times23\times22\times21!}{21!}=25\times24\times23\times22=303,600$