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Three pairs of siblings, each pair consisting of one girl

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Director
Joined: 22 Mar 2011
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Three pairs of siblings, each pair consisting of one girl [#permalink]  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

_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Math Expert
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Posts: 23355
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Kudos [?]: 28717 [7] , given: 2821

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]  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.

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Director
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
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Kudos [?]: 529 [0], given: 43

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]  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.

By far the fastest and most elegant solution!

Those who want to play with combinatorics are invited to provide an alternate solution.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Director
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 73

Kudos [?]: 529 [2] , given: 43

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]  05 Aug 2012, 23:23
2
KUDOS
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.

_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Manager
Joined: 28 Feb 2012
Posts: 115
GPA: 3.9
WE: Marketing (Other)
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Kudos [?]: 20 [0], given: 17

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]  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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If you found my post useful and/or interesting - you are welcome to give kudos!

Manager
Joined: 12 Oct 2011
Posts: 133
GMAT 1: 700 Q48 V37
GMAT 2: 720 Q48 V40
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Kudos [?]: 66 [0], given: 23

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]  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.

Aren't there only 5! total arrangements around a table for 6 people?
Math Expert
Joined: 02 Sep 2009
Posts: 23355
Followers: 3604

Kudos [?]: 28717 [1] , given: 2821

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]  07 Nov 2012, 04:27
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Expert's post
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.

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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Manager
Joined: 25 Jun 2012
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Kudos [?]: 41 [0], given: 15

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]  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.

Hi, couldnt understand why to devide by 6! ??
Intern
Joined: 07 Mar 2013
Posts: 35
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Kudos [?]: 3 [0], given: 80

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]  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.

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
Senior Manager
Joined: 07 Apr 2012
Posts: 465
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Kudos [?]: 8 [0], given: 58

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]  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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