To solve this problem, we need to find the number of ways the students can be seated given the constraints of the problem.
1. We have a school bus with 25 seats, arranged in 5 rows of 5 seats each.
2. There are 15 students from the first grade and 5 students from the second grade.
3. All the second-grade students must sit in the first row, which has exactly 5 seats.
### Step-by-Step Solution:
#### Step 1: Seating the Second-Grade Students
Since all 5 second-grade students must sit in the first row, we need to arrange these 5 students in 5 seats. The number of ways to arrange 5 students in 5 seats (a permutation) is calculated as [tex]\( 5P5 \)[/tex].
[tex]\[ 5P5 = \frac{5!}{(5-5)!} = 5! \][/tex]
#### Step 2: Seating the First-Grade Students
After seating the second-grade students, there are 20 seats left (since the first row is fully occupied). Now, we need to arrange 15 first-grade students in these 20 remaining seats. The number of ways to arrange 15 students in 20 seats (a permutation) is calculated as [tex]\( 20P15 \)[/tex].
[tex]\[ 20P15 = \frac{20!}{(20-15)!} = \frac{20!}{5!} \][/tex]
#### Step 3: Combining Both Arrangements
To find the total number of ways to seat all the students in the bus, we need to multiply the number of ways to arrange the second-grade students with the number of ways to arrange the first-grade students.
[tex]\[ 5P5 \times 20P15 = 5! \times \frac{20!}{5!} \][/tex]
Simplifying and calculating:
[tex]\[ 5P5 = 120 \][/tex]
[tex]\[ 20P15 = \frac{20!}{5!} = 2432902008176640000 \][/tex]
Therefore, the total number of arrangements is:
[tex]\[
5P5 \times 20P15 = 120 \times 2432902008176640000 = 2432902008176640000
\][/tex]
#### Conclusion
The correct option is B:
[tex]\[
5P5 \times 20P15
\][/tex]
So, the answer is option B: [tex]\( \boxed{5P5 \times 20P15} \)[/tex].
