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

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Manager
Joined: 25 Nov 2011
Posts: 193
Location: India
Concentration: Technology, General Management
GPA: 3.95
WE: Information Technology (Computer Software)
5 girls and 3 boys are arranged randomly in a row  [#permalink]

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19 Feb 2012, 05:08
2
15
00:00

Difficulty:

65% (hard)

Question Stats:

64% (02:37) correct 36% (02:22) wrong based on 342 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

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-Aravind Chembeti

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Joined: 02 Sep 2009
Posts: 50627
Re: 5 girls and 3 boys are arranged randomly in a row  [#permalink]

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19 Feb 2012, 05:25
7
8
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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Intern
Joined: 07 Feb 2013
Posts: 1
Re: 5 girls and 3 boys are arranged randomly in a row  [#permalink]

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06 Mar 2013, 18:28
1
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: 50627
Re: 5 girls and 3 boys are arranged randomly in a row  [#permalink]

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07 Mar 2013, 01:01
1
1
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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Senior Manager
Joined: 23 Oct 2010
Posts: 353
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Re: 5 girls and 3 boys are arranged randomly in a row  [#permalink]

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09 Mar 2013, 02:17
1
1
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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Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Manager
Joined: 02 Sep 2012
Posts: 208
Location: United States
Concentration: Entrepreneurship, Finance
GMAT Date: 07-25-2013
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Re: 5 girls and 3 boys are arranged randomly in a row  [#permalink]

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09 Mar 2013, 05:06
bunnel,
Any problem of similar type available in this forum??? if so can you please tag those ..thanks in advance
Math Expert
Joined: 02 Sep 2009
Posts: 50627
5 girls and 3 boys are arranged randomly in a row  [#permalink]

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10 Mar 2013, 05:02
2
4
Intern
Joined: 03 Aug 2017
Posts: 1
Re: 5 girls and 3 boys are arranged randomly in a row  [#permalink]

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02 Dec 2017, 09:10
Bunuel wrote:
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;

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

Can you please show the math for $$C^2_3$$?
Math Expert
Joined: 02 Sep 2009
Posts: 50627
Re: 5 girls and 3 boys are arranged randomly in a row  [#permalink]

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02 Dec 2017, 09:26
Bunuel wrote:
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;

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

Can you please show the math for $$C^2_3$$?

$$3C2 = \frac{3!}{2!*(3-2)!}$$

21. Combinatorics/Counting Methods

For more:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
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Re: 5 girls and 3 boys are arranged randomly in a row &nbs [#permalink] 02 Dec 2017, 09:26
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