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Ten students are waiting to board a school bus. "Student 1" wants to

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Bunuel
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Ten students are waiting to board a school bus. "Student 1" wants to [#permalink] New post   29 Dec 2021, 04:15
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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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Regor60
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Ten students are waiting to board a school bus. "Student 1" wants to [#permalink] New post   Updated on: 02 Jan 2022, 07:02
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There are 6 ways the 3 students could enter the bus with no constraints.

Given the constraints, there are 2 ways for them to enter:

132 and 312

So, 1/3 of the orderings are possible.

10!/3 = 1209600

This assumes that order matters, that the students are unique.

Students "preference" doesn't clarify this ambiguity.


Answers should be recast as factorials. No way GMAT actually would require all the multiplying.


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Originally posted by Regor60 on 30 Dec 2021, 14:45.
Last edited by Regor60 on 02 Jan 2022, 07:02, edited 1 time in total.
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Chandanasreenivas
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Ten students are waiting to board a school bus. "Student 1" wants to [#permalink] New post   01 Jan 2022, 22:05
Since the 3 students want to board in order, we need to arrange the other 7 around them.

__ I __ III __ II __
a + b + c + d = 7

Here a, b, c and d can be any numbers from 0-7.

Therefore the total ways in which the students can board will be 10C3 - 120

Sine the 3 students can also board as __III__I__II__, the students can assign themselves in 120 ways in this way as well.

Also, the 7 students can arrange themselves in 7! ways.

So the total will be = (120+120)*7! = 1209600
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Lodz697
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Re: Ten students are waiting to board a school bus. "Student 1" wants to [#permalink] New post   02 Apr 2024, 04:45
Hello Bunuel!

Could you explain the correct procedure for this question?

Thanks in advance!
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Bunuel
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Re: Ten students are waiting to board a school bus. "Student 1" wants to [#permalink] New post   04 Apr 2024, 00:57
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Lodz697 wrote:
Hello Bunuel!

Could you explain the correct procedure for this question?

Thanks in advance!

­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

10 students can be arranged in 10! ways. The three students mentioned can be arranged in 3! = 6 ways. However, only 1/3 of these arrangements will meet the constraints mentioned: ...1...3...2... and ...3...1...2... Therefore, only 1/3 of all the arrangements of boarding will have the required lineup. Hence, the answer is 10!/3.

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Hope it helps.­
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Re: Ten students are waiting to board a school bus. "Student 1" wants to [#permalink]
04 Apr 2024, 00:57
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