Question: There are eight teams in a league. In how many ways can the teams finish first, second, and third? (Assume there are no ties.)
1. **State the problem:** We want to find the number of ways to arrange 8 teams in the first, second, and third positions without any ties.
2. **Formula used:** This is a permutation problem where order matters. The number of ways to arrange $k$ items out of $n$ is given by the permutation formula:
$$P(n,k) = \frac{n!}{(n-k)!}$$
3. **Apply the formula:** Here, $n=8$ teams and $k=3$ positions.
$$P(8,3) = \frac{8!}{(8-3)!} = \frac{8!}{5!}$$
4. **Calculate factorials:**
$$8! = 8 \times 7 \times 6 \times 5!$$
So,
$$P(8,3) = \frac{8 \times 7 \times 6 \times 5!}{5!}$$
5. **Cancel common factors:**
$$P(8,3) = 8 \times 7 \times 6$$
6. **Multiply:**
$$8 \times 7 = 56$$
$$56 \times 6 = 336$$
7. **Final answer:** There are $336$ ways for the teams to finish first, second, and third with no ties.