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The problem asks how many possible ways there are to place the runners in first, second, and third positions in a race with 18 participants.

This is a **permutation** problem because the order in which we place the runners matters (i.e., first, second, and third are distinct positions). The formula for permutations when choosing \( r \) items from \( n \) options is:

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

Where:
- \( n \) is the total number of participants (18 in this case),
- \( r \) is the number of positions to fill (3 for first, second, and third place).

For this problem:

\[
P(18, 3) = \frac{18!}{(18 - 3)!} = \frac{18!}{15!}
\]

This simplifies to:

\[
P(18, 3) = 18 \times 17 \times 16 = 4896
\]

So, there are **4,896** possible ways to place the runners in first, second, and third place.

Would you like further clarification, or do you have any additional questions? Here are some related questions you might consider:

1. What would the number of ways be if we were placing the top 5 runners instead of 3?
2. How does the formula change if the order didn't matter (combination)?
3. If only the first two positions mattered, how many different arrangements would there be?
4. How would the number of possibilities change if two runners had to tie for first place?
5. What is the probability that a specific runner finishes in first place?

**Tip:** In permutation problems, the order of selection matters, whereas in combination problems, it doesn't.

There are 18 people running in a cross country race. How many possible ways are there to place the runners in first, second, and third?

Learn how to calculate the number of possible ways to place 18 runners in first, second, and third positions in a cross country race using permutations. This problem covers key combinatorics concepts and is suitable for high school students in Grades 10-12.

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