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5 girls and 3 boys are arranged randomly in a row

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Chembeti
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5 girls and 3 boys are arranged randomly in a row [#permalink] New post   19 Feb 2012, 06:08
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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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Bunuel
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink] New post   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;

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

Answer: A.
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Bunuel
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5 girls and 3 boys are arranged randomly in a row [#permalink] New post   10 Mar 2013, 06:02
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skamal7 wrote:
bunnel,
Any problem of similar type available in this forum??? if so can you please tag those ..thanks in advance


Here you go:
how-many-different-arrangements-of-a-b-c-d-and-e-are-pos-106764.html
a-group-of-four-women-and-three-men-have-tickets-for-seven-a-88604.html
two-couples-and-one-single-person-are-seated-at-random-in-a-92400.html
in-how-many-different-ways-can-4-ladies-and-4-gentlemen-be-102187.html
4-couples-are-seating-at-a-round-tables-how-many-ways-can-131048.html
at-a-party-5-people-are-to-be-seated-around-a-circular-104101.html
seven-family-members-are-seated-around-their-circular-dinner-102184.html
seven-men-and-seven-women-have-to-sit-around-a-circular-11473.html
a-group-of-8-friends-sit-together-in-a-circle-alice-betty-106928.html
seven-men-and-five-women-have-to-sit-around-a-circular-table-98185.html
a-group-of-8-friends-sit-together-in-a-circle-alice-betty-106928.html
find-the-number-of-ways-in-which-four-men-two-women-and-a-106919.html
the-coen-family-consists-of-a-father-a-mother-two-children-133489.html
in-how-many-ways-can-you-sit-8-people-on-a-bench-if-3-of-128955.html
on-how-many-ways-can-the-letters-of-the-word-computer-be-a-47764.html
in-how-many-ways-can-6-people-a-b-c-d-e-f-be-seated-127234.html
two-couples-and-one-single-person-are-seated-at-random-in-a-125807.html
what-fraction-of-seven-lettered-words-formed-using-the-141789.html
6-persons-are-going-to-theater-and-will-sit-next-to-each-oth-95974.html
find-the-number-of-ways-in-which-four-men-two-women-and-a-106919.html
seven-family-members-are-seated-around-their-circular-dinner-102184.html
in-how-many-different-ways-can-4-ladies-and-4-gentlemen-be-102187.html
7-people-round-table-108872.html
possible-arrangements-for-the-word-review-if-one-e-can-t-be-90174.html
suppose-that-n-people-are-seated-in-a-single-row-of-n-theate-101390.html
a-group-of-8-friends-sit-together-in-a-circle-alice-betty-106928.html

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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink] New post   06 Mar 2013, 19:28
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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!
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink] New post   07 Mar 2013, 02:01
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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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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink] New post   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] New post   09 Mar 2013, 06:06
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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] New post   02 Dec 2017, 10:10
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;

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

Answer: A.


Can you please show the math for \(C^2_3\)?
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink] New post   02 Dec 2017, 10:26
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pharmademic wrote:
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;

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

Answer: A.


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 ! ! !
Ultimate GMAT Quantitative Megathread


Hope it helps.
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink] New post   16 Jul 2019, 16:55
We have restrictions only for the seats that are at the end. It doesn't matter who sits on the seats which are not at the end.

Number of ways students can sit on the end seats without any constraints= 8*7=56

Number of ways students can sit on the end seats, when there is one boy on each end= 3*2=6
Probability that there is one boy on each end= 6/56=3/28

Number of ways students can sit on the end seats, when there is one girl on each end = 5*4=20
Probability that there is one girl on each end= 20/56=5/14





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
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink] New post   03 Oct 2020, 03:23
Another approach using FCP -
Total arrangements = 8!
Arranging 6 people in between 2 boys/ girls = 6!
1. B - 6! - B (Format)

3 * 2 * 6!/ 8! = 3/28

2. G - 6! - G
5*4*6! = 5/14
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink] New post   04 Oct 2020, 02:23
Total possible cases with 5 girls and 3 boys = 8!

B _ _ _ _ _ _ B

So to select 2 boys for each end and these boys can be at either ends so = 3C2 * 2! = 6
AND possible combination for middle seats i.e for 1 boy and 5 girls = total 6!

Probability = Boys at either ends / Total possibilities = 6 * 6! / 8! = 3/28

Similarly

G _ _ _ _ _ _ G

So to select 2 girls for each end and these girls can be at either end so = 5C2 * 2! = 20
AND possible combination for middle seats i.e for 3 boys and 3 girls = total 6!

Probability = Girls at either ends / Total possibilities = 20 * 6! / 8! = 5/14
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink] New post   04 Oct 2020, 03:22
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Girls: 5 and Boys: 3

Total: 8 => Total number of ways to arrange 8 people is 8!

One boy at each end: Select 2 boys out of 3: \(^3{C_2}\). Now, these two can be arranged in 2! ways. The remaining 5 girls and 1 boy can be arranged in 6! ways.

=> \(^3{C_2}\) * 2! * 6!

=> Probability: \(\frac{(^3{C_2} * 2! * 6!) }{ 8!}\) = \(\frac{3 }{28}\)


One girl at each end: Select 2 girls out of 5: \(^5{C_2}\). Now, these two can be arranged in 2! ways. The remaining 3 girls and 3 boys can be arranged in 6! ways.

=> \(^5{C_2}\) * 2! * 6!

=> Probability: \(\frac{(^5{C_2} * 2! * 6!) }{ 8!}\) = \(\frac{5 }{14}\)

Answer A
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink] New post   02 Aug 2021, 11:13
My approach:

1) B------B --> (6!/5!) / (8!/3!5!) = 3/28

2) G------G --> (6!/3!3!) / (8!/5!3!) = 5/14

Answer A
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Re: 5 girls and 3 boys are arranged randomly in a row [#permalink] New post   18 May 2022, 02:09
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;

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

Answer: A.

Alternatively, since there is not need for boys and girls to be unique:
if you put 2 boys at ends, 1 boy and 5 girls are left to distribute among 6 slots, 6C1=6. and 5 girls and 3 boys between 8 places is 8C3=56. => 6/56/=3/28
if you put 2 girls at ends, 3 boys and 3 girls are left to distribute among 6 slots, 6C3=20. and 5 girls and 3 boys between 8 places is 8C3=56. => 20/56/=5/14
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5 girls and 3 boys are arranged randomly in a row [#permalink] New post   16 Sep 2023, 00:27
Given: 5 girls and 3 boys are arranged randomly in a row.
Asked: Find the probability that:

1) there is one boy on each end.
Number of ways to arrange 5 girls and 3 boys such that there is one boy on each end = 3C2*2*6!
Total number of ways to arrange 5 girls and 3 boys = 8!
The probability that there is one boy on each end = 3C2*2*6!/8! = 3/28

2) There is one girl on each end.
Number of ways to arrange 5 girls and 3 boys such that there is one girl on each end = 5C2*2*6!
Total number of ways to arrange 5 girls and 3 boys = 8!
The probability that there is one boy on each end = 5C2*2*6!/8! = 5/14

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