Seating Arrangements in a School Bus
The number of ways students can be seated in a school bus depends on several factors:
- The number of seats \(n\) in the bus
- The number of students \(r\)
- Whether the order of seating matters (permutation) or not (combination)
- Whether there are more students than seats or vice versa
Case 1: Exactly as many students as seats (\(n = r\))
Let’s assume there are 10 students and 10 seats. The task is to find the number of ways the students can be seated.
Since there are exactly as many students as seats, we use the permutation formula:
\[
P(n, r) = r! = 10!
\]
Calculating \(10!\):
\[
10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800
\]
Therefore, the students can be seated in 3,628,800 ways.
Case 2: More seats than students (\(n > r\))
Let’s assume there are 15 seats and 10 students. We want to find the number of ways to seat 10 students in 15 seats.
We use the permutation formula:
\[
P(n, r) = \frac{n!}{(n - r)!}
\]
For \(n = 15\) seats and \(r = 10\) students:
\[
P(15, 10) = \frac{15!}{(15 - 10)!} = \frac{15!}{5!}
\]
Let's calculate this:
\[
15! = 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6
\]
\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\]
So:
\[
P(15, 10) = \frac{15!}{5!} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6}{120}
\]
This equals:
\[
P(15, 10) = 1,816,214,400
\]
Therefore, there are 1,816,214,400 ways to seat 10 students in 15 seats.
Case 3: More students than seats (\(n < r\))
Let’s assume there are 8 seats and 10 students. We need to choose 8 students from 10 and then arrange them in the seats.
This is a two-step process:
- First, choose which 8 students will sit from the 10. This is a combination:
\[
\binom{r}{n} = \binom{10}{8} = \frac{10!}{8!(10-8)!} = \frac{10!}{8!2!}
\]
Simplifying:
\[
\binom{10}{8} = \frac{10 \times 9}{2 \times 1} = 45
\]
- Now arrange the chosen 8 students in the 8 seats, which is:
\[
8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320
\]
Therefore, the total number of seating arrangements is:
\[
\text{Total ways} = 45 \times 8! = 45 \times 40,320 = 1,814,400
\]
So, there are
1,814,400 ways to seat 8 students chosen from 10.