In this scenario, we need to calculate how many different ways the first three positions can be filled from a total of 19 horses. Since the order in which the horses finish matters, this is a permutation problem.

The number of permutations of 19 horses taken 3 at a time is given by the formula for permutations:

$P(n, r) = \frac{n!}{(n - r)!}$

Where:

• $n$ is the total number of horses (19),
• $r$ is the number of positions (3).

Substituting into the formula:

$P(19, 3) = \frac{19!}{(19 - 3)!} = \frac{19!}{16!}$

This simplifies to:

$P(19, 3) = 19 \times 18 \times 17 = 5814$

So, there are 5,814 different possible finishes among the first 3 places in the race.

Would you like more details on permutations, or have any questions?

Here are some related questions to explore:

1. What is the difference between permutations and combinations?
2. How would the result change if only 2 places were considered?
3. How can you calculate the number of ways to choose 3 horses without regard to the order?
4. How does the permutation formula change when all horses have an equal chance of tying?
5. How many different finishes would be possible with 10 horses?

Tip: Always check if the order matters to decide between permutations and combinations.