1. **State the problem:** There are 10 contestants in a speech contest, and we want to find how many possible ways the first, second, and third presenters can be chosen from these contestants. 2. **Concept applied:** This is a permutation problem because the order in which the presenters are chosen matters (first, second, third). 3. **Formula for permutations:** The number of ways to arrange $k$ items out of $n$ is given by $$P(n,k) = \frac{n!}{(n-k)!}$$ where $n!$ is the factorial of $n$. 4. **Apply the formula:** Here, $n=10$ and $k=3$, so $$P(10,3) = \frac{10!}{(10-3)!} = \frac{10!}{7!}$$ 5. **Calculate factorials:** $$10! = 10 \times 9 \times 8 \times 7!$$ So, $$P(10,3) = \frac{10 \times 9 \times 8 \times 7!}{7!}$$ 6. **Cancel common factors:** $$P(10,3) = 10 \times 9 \times 8$$ 7. **Multiply:** $$10 \times 9 = 90$$ $$90 \times 8 = 720$$ 8. **Final answer:** There are **720** possible ways to pick the first, second, and third presenters from 10 contestants.