1. **Problem statement:** We need to find how many different ways 3 students can be chosen from a group of 10 to sit in the front row. 2. **Formula used:** Since the order in which the students are chosen matters for seating, this is a permutation problem. The number of ways to choose and arrange $k$ students from $n$ is given by the permutation formula: $$P(n,k) = \frac{n!}{(n-k)!}$$ 3. **Apply the formula:** Here, $n=10$ and $k=3$. $$P(10,3) = \frac{10!}{(10-3)!} = \frac{10!}{7!}$$ 4. **Calculate factorial values:** $$10! = 10 \times 9 \times 8 \times 7!$$ So, $$P(10,3) = \frac{10 \times 9 \times 8 \times 7!}{7!} = 10 \times 9 \times 8$$ 5. **Final calculation:** $$10 \times 9 = 90$$ $$90 \times 8 = 720$$ 6. **Answer:** There are **720** different ways to choose and arrange 3 students from 10 to sit in the front row.