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Hovkial
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Ten students are waiting to board a school bus. "Student 1" 01 Oct 2019, 10:36
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

42% (03:19) correct 58% (02:37) wrong based on 24 sessions

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Ten students are waiting to board a school bus. "Student 1" wants to board before "Student 2" boards and "Student 2" wants to board after "Student 3" boards. The remaining students have no boarding preferences. What is the total number of ways in which all the students can board the bus?

(A) 120

(B) 720

(C) 1209600

(D) 1814400

(E) 3026640

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Ten students are waiting to board a school bus. "Student 1" 01 Oct 2019, 11:18
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Hi, in my mind the answer is not listed, could that be? 10C3 are the different ways we could arange the 3 students in the 10 slots with the desired positions (1<2<3). For each arrangement, 7! is the way the other students can be distributed. Therefore the answer should be:

10C3 x 7! = 10!/3! = 604,800
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Ten students are waiting to board a school bus. "Student 1" 04 Oct 2019, 23:03
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Hovkial
Ten students are waiting to board a school bus. "Student 1" wants to board before "Student 2" boards and "Student 2" wants to board after "Student 3" boards. The remaining students have no boarding preferences. What is the total number of ways in which all the students can board the bus?

(A) 120

(B) 720

(C) 1209600

(D) 1814400

(E) 3026640
Show more

Can someone please point out what I am missing.

1. 7 students can be arranged in 7! ways.
2. There are 2 possible arrangements for the three students 132 and 312.
There are 8 places where they can go now. Choose 3 by 8C3.
Arrange them by 2.

So, 7! * 8C3 * 2 = 564480
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Ten students are waiting to board a school bus. "Student 1" 05 Oct 2019, 04:32
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navderm


Can someone please point out what I am missing.

1. 7 students can be arranged in 7! ways.
2. There are 2 possible arrangements for the three students 132 and 312.
There are 8 places where they can go now. Choose 3 by 8C3.
Arrange them by 2.

So, 7! * 8C3 * 2 = 564480
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I think you are making an assumption that the three students need to get in immediately next to each other, other students can get into the bus in between S1-S3, S3-S2 or S1-S2.

Overall, student 2 can take any place starting from 3 to 10
Given below are S2 positions -> corresponding combinations to seat S1 & S3 (Multiplied by 2 is used because their positions are interchangeable)
S2=3 -> S1, S3 = 2C2 * 2 = 2
S2=4 -> S1, S3 = 3C2 * 2 = 6
S2=5 -> S1, S3 = 4C2 * 2 = 12
S2=6 -> S1, S3 = 5C2 * 2 = 20
S2=7 -> S1, S3 = 6C2 * 2 = 30
S2=8 -> S1, S3 = 7C2 * 2 = 42
S2=9 -> S1, S3 = 8C2 * 2 = 56
S2=10 -> S1, S3 = 9C2 * 2 = 72

Total positions for S1, S3 = 240
in each of these 240 positions, other 7 kids can be seated in 7! ways
240*7! = 12,09,600

That's option C

Of course, doesn't look like one can solve this within 2 minutes. :-)

Let me know if you have a different approach. :-)
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Ten students are waiting to board a school bus. "Student 1" 05 Oct 2019, 09:37
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AravindSastry
navderm


Can someone please point out what I am missing.

1. 7 students can be arranged in 7! ways.
2. There are 2 possible arrangements for the three students 132 and 312.
There are 8 places where they can go now. Choose 3 by 8C3.
Arrange them by 2.

So, 7! * 8C3 * 2 = 564480

I think you are making an assumption that the three students need to get in immediately next to each other, other students can get into the bus in between S1-S3, S3-S2 or S1-S2.

Overall, student 2 can take any place starting from 3 to 10
Given below are S2 positions -> corresponding combinations to seat S1 & S3 (Multiplied by 2 is used because their positions are interchangeable)
S2=3 -> S1, S3 = 2C2 * 2 = 2
S2=4 -> S1, S3 = 3C2 * 2 = 6
S2=5 -> S1, S3 = 4C2 * 2 = 12
S2=6 -> S1, S3 = 5C2 * 2 = 20
S2=7 -> S1, S3 = 6C2 * 2 = 30
S2=8 -> S1, S3 = 7C2 * 2 = 42
S2=9 -> S1, S3 = 8C2 * 2 = 56
S2=10 -> S1, S3 = 9C2 * 2 = 72

Total positions for S1, S3 = 240
in each of these 240 positions, other 7 kids can be seated in 7! ways
240*7! = 12,09,600

That's option C

Of course, doesn't look like one can solve this within 2 minutes. :-)

Let me know if you have a different approach. :-)
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This is a wonderful explanation.. deserve a kudos

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Hovkial
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Ten students are waiting to board a school bus. "Student 1" 05 Oct 2019, 15:24
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Hovkial
Ten students are waiting to board a school bus. "Student 1" wants to board before "Student 2" boards and "Student 2" wants to board after "Student 3" boards. The remaining students have no boarding preferences. What is the total number of ways in which all the students can board the bus?

(A) 120

(B) 720

(C) 1209600

(D) 1814400

(E) 3026640
Show more

Three (3) places can be chosen out of 10 places in 10C3 ways.

There are only two (2) possible arrangements: S1S3S2 or S3S1S2

The remaining seven (7) places can be arranged in 7! ways.

Therefore, the total number of ways in which boarding can occur equals the product:

10C3 X 2 X 7! = 1209600

ANSWER: (C)

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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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