1. **Problem statement:** We want to find the number of ways to arrange 8 students in a line such that two specific students, a and b, are not standing next to each other. 2. **Total arrangements without restriction:** The total number of ways to arrange 8 distinct students in a line is given by the factorial of 8: $$8! = 40320$$ 3. **Arrangements where a and b are together:** Treat a and b as a single unit. Then we have this unit plus the other 6 students, making 7 units total. The number of ways to arrange these 7 units is: $$7! = 5040$$ Since a and b can be arranged within their unit in 2! ways (a then b or b then a), the total number of arrangements with a and b together is: $$2! \times 7! = 2 \times 5040 = 10080$$ 4. **Arrangements where a and b are NOT together:** Subtract the number of arrangements where a and b are together from the total arrangements: $$8! - 2! \times 7! = 40320 - 10080 = 30240$$ 5. **Answer:** There are **30240** ways to arrange 8 students in a line so that students a and b are not next to each other.