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# The total number of ways in which 10 students can be arrange

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The total number of ways in which 10 students can be arrange [#permalink]  16 Jun 2017, 02:28
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The total number of ways in which 10 students can be arranged in a row such that A is always ahead of B?

a. 2x10!
b. 10! /2
c. 10! x 8!
d. none
[Reveal] Spoiler: OA

Last edited by Carcass on 16 Jun 2017, 06:35, edited 1 time in total.
Edited the question
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Joined: 07 Jun 2014
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Re: The total number of ways in which 10 students can be arrange [#permalink]  16 Jun 2017, 17:23
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Expert's post
Hi 3Newton,

This is probably not a GRE question. Could you share the source?

Let us assume that there are 10 students and 10 places. We can break down the problem into 3 parts:

1. Place student A in any position.
2. Find number of options for student B
3. Arrange remaining 8 students

So let student A in position 1:

Number of options for student B = 9 (every other position is open)

Remaining 8 students: 8!

So total number of combination = 9 \times 8!.

Let student B in position 2:

Number of options for student B = 8 (every other position is open except the position 1 and position 2 is occupied by 2)

Remaining 8 students: 8!

So total number of combination = 8 \times 8!.

Now repeat this for all postions till student A in postion 9 and student B is 10.

Hence summing up all the options = 9 \times 8! + 8 \times 8! ....... 1 \times 8! = 45 \times 8! = 5\times 9 \times 8! = 5 \times 9! = \frac{10!}{2}.

Hence option B is correct!
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Re: The total number of ways in which 10 students can be arrange [#permalink]  16 Jun 2017, 22:43
Hi Sandy,

Thanks for responding.

This is from some book on counting and probability sold in India. How to know what is a GRE question and which ones are not?

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Re: The total number of ways in which 10 students can be arrange [#permalink]  18 Jun 2017, 07:10
Expert's post
Use the satandard GRE books. Like the ones dicussed in GRE Books forum. Like: Manhattan, Kaplan etc.

The reson for the same is that GRE has a very specific set of questions. Its not an all sweeping math aptitude test. If you stick to the format you are likely to score better.
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Re: The total number of ways in which 10 students can be arrange [#permalink]  20 Jun 2017, 08:33
1
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Expert's post
3Newton wrote:
The total number of ways in which 10 students can be arranged in a row such that A is always ahead of B?

a. 2x10!
b. 10! /2
c. 10! x 8!
d. none

Another approach....

We can arrange 10 students in 10! ways.
For HALF of these 10! arrangements, A is ahead of B, and for the other HALF of these 10! arrangements, B is ahead of A
So, 10!/2 = the number of arrangements where A is ahead of B.

[Reveal] Spoiler:
B

Cheers,
Brent
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Brent Hanneson – Founder of greenlighttestprep.com

Check out the online reviews of our course

SVP
Joined: 07 Jun 2014
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GRE 1: 323 Q167 V156
WE: Business Development (Energy and Utilities)
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Re: The total number of ways in which 10 students can be arrange [#permalink]  20 Jun 2017, 08:51
Expert's post
GreenlightTestPrep wrote:
3Newton wrote:
The total number of ways in which 10 students can be arranged in a row such that A is always ahead of B?

a. 2x10!
b. 10! /2
c. 10! x 8!
d. none

Another approach....

We can arrange 10 students in 10! ways.
For HALF of these 10! arrangements, A is ahead of B, and for the other HALF of these 10! arrangements, B is ahead of A
So, 10!/2 = the number of arrangements where A is ahead of B.

[Reveal] Spoiler:
B

Cheers,
Brent

Brilliant!!
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Sandy
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Re: The total number of ways in which 10 students can be arrange   [#permalink] 20 Jun 2017, 08:51
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