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Re M0610
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15 Sep 2014, 23:27
Official Solution:If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other? A. \(\frac{1}{33}\) B. \(\frac{2}{33}\) C. \(\frac{1}{3}\) D. \(\frac{16}{33}\) E. \(\frac{11}{12}\) The probability is \(\frac{16}{33}\). Once we select the first person, then there is \(\frac{1}{11}\) chance that the second person selected is their spouse and \(\frac{10}{11}\) chance that the second person selected is not their spouse. For the third person, there is a \(\frac{8}{10}\) chance that the third is not a spouse of the first two, and for the fourth person, there is a \(\frac{6}{9}\) probability that the fourth is not married to any of the first three. \((\frac{10}{11})*(\frac{8}{10})*(\frac{6}{9})=\frac{16}{33}\) Alternative explanation: \(\frac{2*2*2*2*C_6^4}{C_{12}^4}=\frac{16}{33}\) \(C_6^4\)  represents four couples from the given 6. Basically, we have to take 4 couples out of 6, which can be done in \(C_6^4\) ways, and then from each couple you take one person, which can be done in 2 ways for each couple. With 4 couples, this is \(2^4=2*2*2*2\) ways. One more approach: Say we have couples like this (Aman, Bwoman): A1 and B1 A2 and B2 A3 and B3 A4 and B4 A5 and B5 A6 and B6 So we have 6 men (6 x A) and 6 women (6 x B) You can pick 4 people in the following ways:  4 men  i.e. \(C_6^4 = \frac{6!}{2!*4!} = 15\)
 3 men and 1 woman  i.e. \(C_6^3 * C_3^1\) (1 of 3 and not of 6 b/c only 3 women are left to avoid 'coupling' chosen men) = \(\frac{6!}{3!*3!} * \frac{3!}{1!*2!} = 60\)
 2 men and 2 women  i.e. \(C_6^2 * C_4^2\) (again 4 women left to avoid 'coupling') = \(\frac{6!}{2!*4!} * \frac{4!}{2!*2!} = 90\)
 1 man and 3 women  i.e. \(C_6^1 * C_5^3 = \frac{6!}{1!*5!} * \frac{5!}{2!*3!} = 60\)
 4 women  i.e. \(C_6^4 = \frac{6!}{2!*4!} = 15\)
So there are \(15+60+90+60+15=240\) possibilities. Divide this by general possibilities of picking 4 people out of 12  i.e. \(C_{12}^4 = \frac{12!}{4!*8!} = 495\). \(\frac{240}{495} = \frac{16}{33}\). Yet another approach: Each couple can send only one "representative" to the committee. We can choose 4 couples (as there should be 4 members) to send that one "representatives" to the committee in \(C^4_6\) # of ways. But these 4 chosen couples can send two persons (either husband or wife): \(2*2*2*2=2^4\). So # of ways to choose 4 people out 6 married couples so that none of them would be married to each other is: \(C^4_6*2^4\). Total # of ways to choose 4 people out of 12 is \(C^4_{12}\). \(P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\) Answer: D
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Re: M0610
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25 Oct 2015, 08:27
I was thinking to solve it in the following way that is not correct. Please do let me know what is wrong in this approach.
Prob: No. of ways to select 4 persons who are not married to each other / No. of ways to select 4 people out of 12.
No. of ways to select 4 persons who are not married to each other: 12 * 10 * 8 * 6 (For the first person to be selected there are 12 options, once he is selected then for the second person there are 10 options and so on)
No. of ways to select 4 people out of 12 = (12 * 11 * 10 * 9) / (4 * 3 * 2) = 11 * 5 * 9
Now, when I divide (12 * 10 * 8 * 6) and (11 * 5 * 9) I don't get the answer. Could you please let me know what is wrong in my approach.



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Re: M0610
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27 Oct 2015, 05:25
Hi, The top 12*10*8*6 is not selection it is arrangement. Bottom should also be arrangement then that is 12!/8!
Else make top selection 12*10*8*6/4! And bottom selection 12!/4!*8!
Imagine four letters abcd isn't it 4*3*2*1 arrangement. If it is 4*3*2*1/4! It is selection. Hope I am clear



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Re: M0610
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27 Oct 2015, 05:26
Hi, The top 12*10*8*6 is not selection it is arrangement. Bottom should also be arrangement then that is 12!/8!
Else make top selection 12*10*8*6/4! And bottom selection 12!/4!*8!
Imagine four letters abcd isn't it 4*3*2*1 arrangement. If it is 4*3*2*1/4! It is selection. Hope I am clear



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Re: M0610
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27 Oct 2015, 16:00
Hi Bunuel
I solved it through a different approach please let me know if there is any flaw in this approach
Total outcomes , say B = 12C4
Now first person can be selected from any 12 people, the next guy would have to be selected then from 10 people (leaving the person who is married to the first person selected), similarly third person can be selected from 8 and fourth person from 6
So 12*10*8*6
Now, as per question all people should be treated as same
So (12*10*8*6) should be divided by 4! (I.e.Total Arrangement / identical items arrangement)
Say A= (12*10*8*6)/4!
Now for probability divide A by B and you get 16/33.



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Re: M0610
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02 Nov 2015, 09:59
Thanks a lot jojojoseph for pointing out my mistake.



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Re: M0610
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20 Jun 2017, 12:10
Bunuel wrote: Official Solution:If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other? A. \(\frac{1}{33}\) B. \(\frac{2}{33}\) C. \(\frac{1}{3}\) D. \(\frac{16}{33}\) E. \(\frac{11}{12}\) The probability is \(\frac{16}{33}\). Once we select the first person, then there is \(\frac{1}{11}\) chance that the second person selected is their spouse and \(\frac{10}{11}\) chance that the second person selected is not their spouse. For the third person, there is a \(\frac{8}{10}\) chance that the third is not a spouse of the first two, and for the fourth person, there is a \(\frac{6}{9}\) probability that the fourth is not married to any of the first three. \((\frac{10}{11})*(\frac{8}{10})*(\frac{6}{9})=\frac{16}{33}\) Alternative explanation: \(\frac{2*2*2*2*C_6^4}{C_{12}^4}=\frac{16}{33}\) \(C_6^4\)  represents four couples from the given 6. Basically, we have to take 4 couples out of 6, which can be done in \(C_6^4\) ways, and then from each couple you take one person, which can be done in 2 ways for each couple. With 4 couples, this is \(2^4=2*2*2*2\) ways. One more approach: Say we have couples like this (Aman, Bwoman): A1 and B1 A2 and B2 A3 and B3 A4 and B4 A5 and B5 A6 and B6 So we have 6 men (6 x A) and 6 women (6 x B) You can pick 4 people in the following ways:  4 men  i.e. \(C_6^4 = \frac{6!}{2!*4!} = 15\)
 3 men and 1 woman  i.e. \(C_6^3 * C_3^1\) (1 of 3 and not of 6 b/c only 3 women are left to avoid 'coupling' chosen men) = \(\frac{6!}{3!*3!} * \frac{3!}{1!*2!} = 60\)
 2 men and 2 women  i.e. \(C_6^2 * C_4^2\) (again 4 women left to avoid 'coupling') = \(\frac{6!}{2!*4!} * \frac{4!}{2!*2!} = 90\)
 1 man and 3 women  i.e. \(C_6^1 * C_5^3 = \frac{6!}{1!*5!} * \frac{5!}{2!*3!} = 60\)
 4 women  i.e. \(C_6^4 = \frac{6!}{2!*4!} = 15\)
So there are \(15+60+90+60+15=240\) possibilities. Divide this by general possibilities of picking 4 people out of 12  i.e. \(C_{12}^4 = \frac{12!}{4!*8!} = 495\). \(\frac{240}{495} = \frac{16}{33}\). Yet another approach: Each couple can send only one "representative" to the committee. We can choose 4 couples (as there should be 4 members) to send that one "representatives" to the committee in \(C^4_6\) # of ways. But these 4 chosen couples can send two persons (either husband or wife): \(2*2*2*2=2^4\). So # of ways to choose 4 people out 6 married couples so that none of them would be married to each other is: \(C^4_6*2^4\). Total # of ways to choose 4 people out of 12 is \(C^4_{12}\). \(P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\) Answer: D Hi, Can you please tell me what's wrong with this approach? P(none are married) = 1  P(at least one married) P(at least one married) = 6C1(Choose one couple out of 6)*10C2(Remaining two can be anyone) / 12C4 = (6C1* 10C2)/12C4 = 6/11 Therefore, P(none are married) = 1  6/11 = 5/11



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Re: M0610
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06 Jul 2017, 09:10
Total ways of Selecting 4 people out of 12 = 12C4
One person out of couple and 1 from rest 5 couples (or 10 people) = 2C1 * 10C3 No of ways of selecting the couple (whether we are selecting 1st one or 2nd one etc.) = 6C1
Solution: (2C1 * 10C3 * 6C1) / 12C4 = 16/33



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Re: M0610
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15 Aug 2017, 00:20
can anyone please tell why is this approach wrong ?
probability that no one is married : (12C1 * 10C1*8C1*6C1)/12C4 no of ways to select 1st person= 12C1 no of ways selecting 2nd person: 10C1  excluding the person selected in the 1 and his/her spouse henceforth 8C1 and 6C1.. selecting 4 people will be 12C4.
the ans is 64/11 which is clearly against the probability rule. but i want to know what is wrong with this approach.



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Alternatively:
P(none) = 1  P(one couple is selected OR two couples are selected) (A) Total ways to select 4 from 12 = 12C4 = 495 (B) Total ways to select 1 couple from 6 = 6C1 x (10C2  5C1) = 6*40 = 240 i.e. imagine a couple is chosen thus {AB,_,_}: the 3rd and 4th slots can accommodate any 2 from 10 remaining members less the 5 ways (5C1) you can select a married pair from those very 10 (C) Total ways to select 2 couples from 6 = 6C2 = 15
So P(none) = 1  (240+15)/495 = 16/33



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Re: M0610
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20 Jun 2018, 09:26
What if somebody is married to two persons at the same time within this group of 12 ?



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Re: M0610
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13 Oct 2018, 06:14
Bunuel wrote: If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?
A. \(\frac{1}{33}\) B. \(\frac{2}{33}\) C. \(\frac{1}{3}\) D. \(\frac{16}{33}\) E. \(\frac{11}{12}\) buneul why cant we do it this way 1  (6c2 * 4c4)/(12c4) first we chose 2 couples from 6 and then 4 of 4...am i missing something thanks



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Re: M0610
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13 Oct 2018, 06:51
sdgmat89 wrote: Bunuel wrote: If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?
A. \(\frac{1}{33}\) B. \(\frac{2}{33}\) C. \(\frac{1}{3}\) D. \(\frac{16}{33}\) E. \(\frac{11}{12}\) buneul why cant we do it this way 1  (6c2 * 4c4)/(12c4) first we chose 2 couples from 6 and then 4 of 4...am i missing something thanks Sorry i am no buneul ! No you can't do that. How do you know that out of a total 12 persons (8 couples) which 4 persons (i.e 2 couples) are couples? For example Suppose there are 2 Green 2 White and 2 Red balls in a box and we have to find the probability that 2 randomly selected balls without replacement is not green? What you propose is to first select green color out of three colors available, i.e 3C1, then two balls of green color 2C2 [13C1*2C2]/6C2. But that is not possible, since all these 6 balls are mixed in the box and we can only pick out balls from the box not the colors. So, first we select 4 NOT green balls out of total 6 balls 6C4 and then divide it by total ways of selecting 2 balls out of 6 i.e 6C2 4C2/6C2Same is the case here. You cannot select couples from a group of 12 persons. Hope this helps !



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Re: M0610
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13 Oct 2018, 07:15
lukachiri wrote: sdgmat89 wrote: Bunuel wrote: If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?
A. \(\frac{1}{33}\) B. \(\frac{2}{33}\) C. \(\frac{1}{3}\) D. \(\frac{16}{33}\) E. \(\frac{11}{12}\) buneul why cant we do it this way 1  (6c2 * 4c4)/(12c4) first we chose 2 couples from 6 and then 4 of 4...am i missing something thanks Sorry i am no buneul ! No you can't do that. How do you know that out of a total 12 persons (8 couples) which 4 persons (i.e 2 couples) are couples? For example Suppose there are 2 Green 2 White and 2 Red balls in a box and we have to find the probability that 2 randomly selected balls without replacement is not green? What you propose is to first select green color out of three colors available, i.e 3C1, then two balls of green color 2C2 [13C1*2C2]/6C2. But that is not possible, since all these 6 balls are mixed in the box and we can only pick out balls from the box not the colors. So, first we select 4 NOT green balls out of total 6 balls 6C4 and then divide it by total ways of selecting 2 balls out of 6 i.e 6C2 4C2/6C2Same is the case here. You cannot select couples from a group of 12 persons. Hope this helps ! Thanks for the explanation.But i dont think its same case. When balls are mixed in bag its different and balls are 2 each and separate its different. Here he has mentioned there are 6 couples. You cannot expect them to be mixed. Couples are spearate and most important you get to chose the couple complete.y Im worried i missed out some logic here. When you say 2 green, 2 red and 2 blue balls are there...and dont pick up green... yu can surely go and chose 2c1 4c2.. its always visible to you only matter of selection is needed. Similarly in couples qs we can chose the couples first and then 4 from 2 couples is fine... em just not getting the point i missed. pl correct me if i am wrong.



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sdgmat89 wrote: Thanks for the explanation.But i dont think its same case.
When balls are mixed in bag its different and balls are 2 each and separate its different.
Here he has mentioned there are 6 couples. You cannot expect them to be mixed. Couples are spearate and most important you get to chose the couple complete.y
Im worried i missed out some logic here.
When you say 2 green, 2 red and 2 blue balls are there...and dont pick up green... yu can surely go and chose 2c1 4c2..
its always visible to you only matter of selection is needed.
Similarly in couples qs we can chose the couples first and then 4 from 2 couples is fine...
em just not getting the point i missed.
pl correct me if i am wrong. Actually that was a similar situation and the best i could come up with, although it can be phrased it a better way. @Buneul can give you a better example. Now, you pointed out something "mentioned there are 6 couples". Question stem says that there is ' a group of 6 married couples' and not just "6 couples". Again, it is another hint on how to approach the question. Had it been only "6 couples", then its open for debate. Hope you get the point here, selecting from 'a group of 6 couples' and selecting from '6 couples' are two different things. Aren't they ?. In the former case you first have to identify the couples from the group.










