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# 5 girls and 3 boys are arranged randomly in a row

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Senior Manager
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5 girls and 3 boys are arranged randomly in a row [#permalink]

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19 Feb 2012, 06:08
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Difficulty:

65% (hard)

Question Stats:

63% (02:51) correct 37% (02:05) wrong based on 224 sessions

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5 girls and 3 boys are arranged randomly in a row. Find the probability that:

1) there is one boy on each end.
2) There is one girl on each end.

A. 3/28 & 5/14
B. 5/28 & 3/14
C. 5/14 & 3/28
D. 3/28 & 1/14
E. 1/28 & 5/14
[Reveal] Spoiler: OA

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-Aravind Chembeti

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Kudos [?]: 106563 [4] , given: 11628

Re: 5 girls and 3 boys are arranged randomly in a row [#permalink]

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19 Feb 2012, 06:25
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Chembeti wrote:
5 girls and 3 boys are arranged randomly in a row. Find the probability that:

1) there is one boy on each end.
2) There is one girl on each end.

A. 3/28 & 5/14
B. 5/28 & 3/14
C. 5/14 & 3/28
D. 3/28 & 1/14
E. 1/28 & 5/14

Total # of arrangements is 8!;

1. The # of arrangements where there is one boy on each end is $$C^2_3*2*6!$$, where $$C^2_3$$ is # of ways to select which 2 boys will be on that positions, *2 # of way to arrange them on that position (A------B and B------A) and 6! is # of arrangements of 6 people left;

$$P=\frac{C^2_3*2*6!}{8!}=\frac{3}{28}$$.

2. The # of arrangements where there is one girl on each end is $$C^2_5*2*7!$$, where $$C^2_5$$ is # of ways to select which 2 girls will be on that positions, *2 # of way to arrange them on that position (A------B and B------A) and 6! is # of arrangements of 6 people left;

$$P=\frac{C^2_5*2*6!}{8!}=\frac{5}{14}$$.

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Kudos [?]: 106563 [2] , given: 11628

5 girls and 3 boys are arranged randomly in a row [#permalink]

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10 Mar 2013, 06:02
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Kudos [?]: 1 [1] , given: 11

Re: 5 girls and 3 boys are arranged randomly in a row [#permalink]

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06 Mar 2013, 19:28
1
KUDOS
Hi Bunel, firstly I would like to thank you for all the help you've given to the forum! You have been nothing but superb!

I have a question on this though, would it be possible also to solve the question as below?
I had solved it like this initially and I'm not sure now if I have some mistakes in my fundamental theory.
It would be great if you could help clarify.
Thank you!

1)
[m]P=(P^2_3*6!)/8!

2)
[m]P=(P^2_5*6!)/8!
Math Expert
Joined: 02 Sep 2009
Posts: 39062
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Kudos [?]: 106563 [1] , given: 11628

Re: 5 girls and 3 boys are arranged randomly in a row [#permalink]

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07 Mar 2013, 02:01
1
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Expert's post
swyw wrote:
Hi Bunel, firstly I would like to thank you for all the help you've given to the forum! You have been nothing but superb!

I have a question on this though, would it be possible also to solve the question as below?
I had solved it like this initially and I'm not sure now if I have some mistakes in my fundamental theory.
It would be great if you could help clarify.
Thank you!

1)
[m]P=(P^2_3*6!)/8!

2)
[m]P=(P^2_5*6!)/8!

Yes, your approach is fine: 3P2=3C2*2! and 5P2=5C2*2!.

Hope it's clear.
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Kudos [?]: 354 [1] , given: 73

Re: 5 girls and 3 boys are arranged randomly in a row [#permalink]

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09 Mar 2013, 03:17
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one boy on each end = (3/8)*(2/7)=3/28
one girl on each end =(5/8)*(4/7)=5/14
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink]

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09 Mar 2013, 06:06
bunnel,
Any problem of similar type available in this forum??? if so can you please tag those ..thanks in advance
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink]

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19 Jul 2014, 09:14
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink]

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24 Aug 2015, 12:08
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink]

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10 Sep 2016, 15:34
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: 5 girls and 3 boys are arranged randomly in a row   [#permalink] 10 Sep 2016, 15:34
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