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5 girls and 3 boys are arranged randomly in a row

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

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New post 19 Feb 2012, 05:08
2
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A
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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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Re: 5 girls and 3 boys are arranged randomly in a row  [#permalink]

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New post 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;

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

Answer: A.
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Posts: 1
Re: 5 girls and 3 boys are arranged randomly in a row  [#permalink]

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

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

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

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

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New post 10 Mar 2013, 05:02
2
4
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

Hope it helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 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;

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

Answer: A.


Can you please show the math for \(C^2_3\)?
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Posts: 50627
Re: 5 girls and 3 boys are arranged randomly in a row  [#permalink]

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New post 02 Dec 2017, 09:26
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



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Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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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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