To solve this question, let's consider the details provided and break it down step by step.
1. Arrangement of First-Grade Students:
- There are 15 first-grade students.
- These 15 students must occupy the first 3 rows of the bus.
- There are 5 seats per row, so the total number of seats in the first 3 rows is [tex]\(3 \times 5 = 15\)[/tex] seats.
- We need to arrange these 15 students in 15 seats.
2. Arrangement of Second-Grade Students:
- There are 5 second-grade students.
- These 5 students will occupy the remaining 2 rows of the bus.
- There are 5 seats per row, so the total number of seats in the remaining 2 rows is [tex]\(2 \times 5 = 10\)[/tex] seats.
- We need to choose 5 out of these 10 seats to accommodate the second-grade students.
Given the principles of permutations and combinations:
- Arranging 15 first-grade students in 15 seats is simply a permutation problem. The number of ways to arrange [tex]\(n\)[/tex] items in [tex]\(n\)[/tex] positions is [tex]\(n!\)[/tex], which is equivalent to [tex]\({}_{15}P_{15}\)[/tex].
- Choosing 5 out of 10 seats for the second-grade students is a combination problem. The number of ways to choose [tex]\(r\)[/tex] items from [tex]\(n\)[/tex] items is [tex]\({}_nC_r\)[/tex], which is [tex]\({}_{10}C_5\)[/tex].
Thus, the total number of ways to seat both the first-grade and second-grade students is given by the product of these two results:
[tex]\[
{}_{15}P_{15} \times {}_{10}C_5
\][/tex]
Given the multiple-choice options:
A. [tex]\({}_{25}P_{20}\)[/tex]
B. [tex]\({}_{5}P_{5} \times {}_{20}P_{15}\)[/tex]
C. [tex]\({}_{15}C_{15} \times {}_{10}C_5\)[/tex]
D. [tex]\({}_{15}P_{15} \times {}_{10}P_5\)[/tex]
E. [tex]\({}_{15}P_{15} \times {}_{10}C_5\)[/tex]
The correct answer is:
E. [tex]\({}_{15}P_{15} \times {}_{10}C_5\)[/tex]
Thus, the total number of ways the students can be seated is the product of [tex]\({}_{15} P _{15}\)[/tex] and [tex]\({}_{10} C _{5}\)[/tex].
The result of this calculation is:
[tex]\[
329533940736000
\][/tex]
