Chapter 4 - Section 4.6 - Counting Rule, Factorials, Combinations, and Permutations - Exercises - Page 168: 4.90

$_{12}P_{3}$ = 1320 There are 1320 permutations.

All horses are different from one another, therefore the order of selections is important; we find the number of permutations or arrangements using the following formula: $12P_{3}$ = $\frac{12!}{(12-3)!}$ $= \frac{12!}{9!} = 1320$ Thus, there are 1320 permutations or arrangements for selecting 3 horses out of 12 horses.