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Three pairs of siblings, each pair consisting of one girl

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Three pairs of siblings, each pair consisting of one girl [#permalink] New post 05 Aug 2012, 03:32
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Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

(A) \, \frac{1}{2}
(B) \, \frac{1}{4}
(C) \, \frac{1}{6}
(D) \, \frac{1}{8}
(E) \, \frac{1}{16}
[Reveal] Spoiler: OA

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Re: Three pairs of siblings, each pair consisting of one girl [#permalink] New post 05 Aug 2012, 04:08
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EvaJager wrote:
Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

(A) \, \frac{1}{2}
(B) \, \frac{1}{4}
(C) \, \frac{1}{6}
(D) \, \frac{1}{8}
(E) \, \frac{1}{16}


Notice that we need a girl to be to the left of her sibling, but not necessarily right to the left of him (meaning that if B and G are siblings, then GB arrangement as well as for example G*B arrangement is possible).

Now, the probability that one particular sibling is seated that way is 1/2 (a girl can be either to the left of her sibling or to the right), the probability that two siblings are seated that way is 1/2*1/2 and the probability that all three siblings are seated that way is 1/2*1/2*1/2=1/8.

Answer: D.
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Re: Three pairs of siblings, each pair consisting of one girl [#permalink] New post 05 Aug 2012, 04:35
Bunuel wrote:
EvaJager wrote:
Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

(A) \, \frac{1}{2}
(B) \, \frac{1}{4}
(C) \, \frac{1}{6}
(D) \, \frac{1}{8}
(E) \, \frac{1}{16}


Notice that we need a girl to be to the left of her sibling, but not necessarily right to the left of him (meaning that if B and G are siblings, then GB arrangement as well as for example G*B arrangement is possible).

Now, the probability that one particular sibling is seated that way is 1/2 (a girl can be either to the left of her sibling or to the right), the probability that two siblings are seated that way is 1/2*1/2 and the probability that all three siblings are seated that way is 1/2*1/2*1/2=1/8.

Answer: D.



By far the fastest and most elegant solution!

Those who want to play with combinatorics are invited to provide an alternate solution.
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Re: Three pairs of siblings, each pair consisting of one girl [#permalink] New post 05 Aug 2012, 23:23
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EvaJager wrote:
Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

(A) \, \frac{1}{2}
(B) \, \frac{1}{4}
(C) \, \frac{1}{6}
(D) \, \frac{1}{8}
(E) \, \frac{1}{16}


Here is a solution using combinatorics:

Place the first pair of siblings - we have 6 possibilities for one of them, and 5 for the other one; this gives 6*5, but we have to divide by 2, as only in half of them, the girls will sit on the left of her brother. So, 6*5/2=15 possibilities.
To place the second pair of siblings - similarly, we have 4*3/2=6 possibilities.
Finally, for the last and third pair - 2*1/2 = 1 possibility.

Therefore, the requested probability is 15*6/6!= 3*5*6/(2*3*4*5*6)=1/(2*4) = 1/8.

Answer D.
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Re: Three pairs of siblings, each pair consisting of one girl [#permalink] New post 06 Aug 2012, 21:45
Thanks Bunuel, you once again showed that in GMAT in most cases it is more logical thinking than doing quants. I have tried this one with different approaches but still could not come up with solution, but after your explanation it seems so easy and i wonder how i could not come up myself.

Thanks!!!
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Re: Three pairs of siblings, each pair consisting of one girl [#permalink] New post 07 Nov 2012, 02:51
EvaJager wrote:
EvaJager wrote:
Therefore, the requested probability is 15*6/6!= 3*5*6/(2*3*4*5*6)=1/(2*4) = 1/8.

Answer D.

Aren't there only 5! total arrangements around a table for 6 people?
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Re: Three pairs of siblings, each pair consisting of one girl [#permalink] New post 07 Nov 2012, 04:27
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BN1989 wrote:
EvaJager wrote:
EvaJager wrote:
Therefore, the requested probability is 15*6/6!= 3*5*6/(2*3*4*5*6)=1/(2*4) = 1/8.

Answer D.

Aren't there only 5! total arrangements around a table for 6 people?


We are not told that these 6 are seated around a table, so we don't have circular arrangement. The question implies that they are seated like in a row.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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: Three pairs of siblings, each pair consisting of one girl [#permalink] New post 08 Nov 2012, 10:06
EvaJager wrote:
EvaJager wrote:
Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

(A) \, \frac{1}{2}
(B) \, \frac{1}{4}
(C) \, \frac{1}{6}
(D) \, \frac{1}{8}
(E) \, \frac{1}{16}


Here is a solution using combinatorics:

Place the first pair of siblings - we have 6 possibilities for one of them, and 5 for the other one; this gives 6*5, but we have to divide by 2, as only in half of them, the girls will sit on the left of her brother. So, 6*5/2=15 possibilities.
To place the second pair of siblings - similarly, we have 4*3/2=6 possibilities.
Finally, for the last and third pair - 2*1/2 = 1 possibility.

Therefore, the requested probability is 15*6/6!= 3*5*6/(2*3*4*5*6)=1/(2*4) = 1/8.

Answer D.


Hi, couldnt understand why to devide by 6! ??
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Re: Three pairs of siblings, each pair consisting of one girl [#permalink] New post 12 Feb 2014, 03:09
Bunuel wrote:
EvaJager wrote:
Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

(A) \, \frac{1}{2}
(B) \, \frac{1}{4}
(C) \, \frac{1}{6}
(D) \, \frac{1}{8}
(E) \, \frac{1}{16}


Notice that we need a girl to be to the left of her sibling, but not necessarily right to the left of him (meaning that if B and G are siblings, then GB arrangement as well as for example G*B arrangement is possible).

Now, the probability that one particular sibling is seated that way is 1/2 (a girl can be either to the left of her sibling or to the right), the probability that two siblings are seated that way is 1/2*1/2 and the probability that all three siblings are seated that way is 1/2*1/2*1/2=1/8.

Answer: D.

Hi Bunuel,

first sibling can be seated in 1/2 ways. but how do we come about the second sibling probability of 1/2 ? I am bit confused here, can you explain please ?

thanks
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Re: Three pairs of siblings, each pair consisting of one girl [#permalink] New post 29 Jun 2014, 06:43
Why is there a division by 6!?
Re: Three pairs of siblings, each pair consisting of one girl   [#permalink] 29 Jun 2014, 06:43
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