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A school bus has 25 seats, with 5 rows of 5 seats. 15 students from the first grade and 5 students from the second grade travel in the bus. How many ways can the students be seated if all the first-grade students occupy the first 3 rows?

A. 25P20
B. 5P5 × 20P15
C. 15C15 × 10C5
D. 15P15 × 10P5
E. 15P15 × 10C5

A school bus has 25 seats, with 5 rows of 5 seats. 15 students from the first grade-example-1

1 Answer

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Answer:

To determine the number of ways the students can be seated if all the first-grade students occupy the first 3 rows of the school bus, we can break down the problem into two steps:

Step 1: Arrange the first-grade students in the first 3 rows

Since there are 15 first-grade students and 15 seats available in the first 3 rows, we can use the permutation formula to calculate the number of ways they can be arranged. The formula for permutations is nPr = n! / (n - r)!, where n is the total number of items and r is the number of items to be arranged.

Using the permutation formula, we have:

15P15 = 15!

= 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

≈ 1.30767 x 10^12

Step 2: Arrange the second-grade students in the remaining seats

After arranging the first-grade students, we are left with 10 seats for the 5 second-grade students. We can use the combination formula to calculate the number of ways they can be arranged. The formula for combinations is nCr = n! / (r!(n - r)!), where n is the total number of items and r is the number of items to be chosen.

Using the combination formula, we have:

10C5 = 10! / (5!(10 - 5)!)

= 10! / (5! x 5!)

= (10 x 9 x 8 x 7 x 6) / (5 x 4 x 3 x 2 x 1)

= 252

Therefore, the total number of ways the students can be seated is the product of the arrangements of the first-grade students and the combinations of the second-grade students:

1.30767 x 10^12 x 252 ≈ 3.29725 x 10^14

Based on the given answer choices, the closest option to this value is:

D. 15P15 * 10P5

Explanation:

User Andrei Bucurei
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