It is currently 20 Oct 2017, 20:46

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Three pairs of siblings, each pair consisting of one girl

Author Message
TAGS:

### Hide Tags

Director
Joined: 22 Mar 2011
Posts: 610

Kudos [?]: 1058 [1], given: 43

WE: Science (Education)
Three pairs of siblings, each pair consisting of one girl [#permalink]

### Show Tags

05 Aug 2012, 04:32
1
KUDOS
26
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

42% (02:05) correct 58% (01:54) wrong based on 309 sessions

### HideShow timer Statistics

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.

Kudos [?]: 1058 [1], given: 43

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129047 [9], given: 12187

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

### Show Tags

05 Aug 2012, 05:08
9
KUDOS
Expert's post
8
This post was
BOOKMARKED
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.

_________________

Kudos [?]: 129047 [9], given: 12187

Director
Joined: 22 Mar 2011
Posts: 610

Kudos [?]: 1058 [0], given: 43

WE: Science (Education)
Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

### Show Tags

05 Aug 2012, 05:35
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.

Kudos [?]: 1058 [0], given: 43

Director
Joined: 22 Mar 2011
Posts: 610

Kudos [?]: 1058 [4], given: 43

WE: Science (Education)
Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

### Show Tags

06 Aug 2012, 00:23
4
KUDOS
2
This post was
BOOKMARKED
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.

Kudos [?]: 1058 [4], given: 43

Manager
Joined: 28 Feb 2012
Posts: 115

Kudos [?]: 52 [0], given: 17

GPA: 3.9
WE: Marketing (Other)
Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

### Show Tags

06 Aug 2012, 22:45
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!!!
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Kudos [?]: 52 [0], given: 17

Manager
Joined: 12 Oct 2011
Posts: 131

Kudos [?]: 251 [0], given: 23

GMAT 1: 700 Q48 V37
GMAT 2: 720 Q48 V40
Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

### Show Tags

07 Nov 2012, 03:51
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?

Kudos [?]: 251 [0], given: 23

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129047 [1], given: 12187

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

### Show Tags

07 Nov 2012, 05:27
1
KUDOS
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.
_________________

Kudos [?]: 129047 [1], given: 12187

Manager
Joined: 25 Jun 2012
Posts: 71

Kudos [?]: 132 [0], given: 21

Location: India
WE: General Management (Energy and Utilities)
Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

### Show Tags

08 Nov 2012, 11:06
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! ??

Kudos [?]: 132 [0], given: 21

Intern
Joined: 07 Mar 2013
Posts: 28

Kudos [?]: 10 [0], given: 80

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

### Show Tags

12 Feb 2014, 04:09
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

Kudos [?]: 10 [0], given: 80

Senior Manager
Joined: 08 Apr 2012
Posts: 446

Kudos [?]: 79 [0], given: 58

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

### Show Tags

29 Jun 2014, 07:43
Why is there a division by 6!?

Kudos [?]: 79 [0], given: 58

Manager
Joined: 03 May 2013
Posts: 76

Kudos [?]: 13 [0], given: 105

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

### Show Tags

05 Aug 2015, 04:21
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 Eva I dont get eather of the solution , by Bunual or by you, please explain Place the first pair of siblings - we have 6 possibilities for one of them, and 5 for the other one; this gives 6*5

Kudos [?]: 13 [0], given: 105

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2675

Kudos [?]: 1725 [1], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Three pairs of siblings, each pair consisting of one girl [#permalink]

### Show Tags

05 Aug 2015, 05:07
1
KUDOS
Expert's post
vipulgoel wrote:
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 Eva I dont get eather of the solution , by Bunual or by you, please explain Place the first pair of siblings - we have 6 possibilities for one of them, and 5 for the other one; this gives 6*5

Let the siblings be : in the order
Girl Boy
A B
C D
E F

Let the arrangement be _ _ _ _ _ _

So if you place lets say A on the first dash (=6 ways you can place A), you only have 5 places to let B go to. Thus you get 6*5. Eva has divided this and other possible arrangements by 2 to account for the fact 50% of the combinations will have AB_ _ _ _ while 50% will be BA_ _ _ _ . Only cases with AB_ _ _ _ type of combinations are allowed. We can safley assume 50% for either cases as there is no case for a 'bias' in these arrangements.

Bunuel has done the same , albeit in a slightly different manner. Probability of any girl sibling sitting to the right of the boy sibling = 50% or 1/2 (same as above)

Final probability = probability of 1st sibling girl to the left of the boy sibling * probability of 2nd sibling girl to the left of the boy sibling *probability of 3rd sibling girl to the left of the boy sibling = 1/2 * 1/2 * 1/2 = 1/8

Probability can be calculated in 2 ways:

Probability = total favorable cases / total cases (which is what Eva has done) or

Probability = probability of case 1* probability of case 2*probability of case 3 etc .... (which is what Bunuel has done).

You can choose whichever method suits you.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1725 [1], given: 792

Manager
Joined: 03 May 2013
Posts: 76

Kudos [?]: 13 [0], given: 105

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

### Show Tags

05 Aug 2015, 05:27
Thanks Now I got it.little more help, where i am wrong in this ...

Let the siblings be : in the order
Girl Boy
A B
C D
E F

Only three cases are available

ABCDEF ....(3!) no of ways three siblings can be arranges like CDABEF(one of the case out of 6) = 6
+
ACEBDF ...3! *3! ( no of ways ACE and BDF can be arranged them self) = 36
+
ABCEDF OR CDAEBF OR EFACBD ( no of ways ne sibling comes extrem left , then remaining two girls then remaining two boys)

3c1(ne subling out of three) *2! (ways two girls arranged among themself ) * 2! (ways 2 boys arranged among themself ) =24

36+24+6/6!

Kudos [?]: 13 [0], given: 105

Manager
Joined: 01 Apr 2015
Posts: 66

Kudos [?]: 12 [0], given: 139

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

### Show Tags

06 Aug 2015, 05:51
Hi Bunuel,

Please let me know if my approach to the problem is correct. I have seen your other solutions where we multiply by 1/2 whenever we have a condition of sitting/ standing only on left or right.

6!*1/8 ==> 90 (favourable outcome) /6! ==> 90/720 ==> 1/8.

So instead of multiplying 1/2 for each pair, i took (1/2)^3, 3 is the number of pairs. I want to know if this approach is correct in case i have to apply to similar questions.

Kudos [?]: 12 [0], given: 139

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1545

Kudos [?]: 837 [0], given: 5

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

### Show Tags

20 Jun 2017, 18:55
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?

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

Let’s call one pair of siblings Abby and John. Since two people can be arranged in 2 ways, the probability is 1/2 that Abby sits to the left of John and 1/2 that she sits to the right of John.
Using that same logic, we see there is a 1/2 x 1/2 x 1/2 = 1/8 probability that all 3 girls are seated to the left of their male siblings.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 837 [0], given: 5

Re: Three pairs of siblings, each pair consisting of one girl   [#permalink] 20 Jun 2017, 18:55
Display posts from previous: Sort by