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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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? 1/33 2/33 1/3 16/33 11/12

Each couple can send only one "representative" to the committee. We can choose 4 couples (as there should be 4 members) to send only 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}.

total ways of selective 4 ppl from 6 married couples = 12C4 Favorable outcome = 12 *10*8*6 ????

The way you are doing is wrong because 12*10*8*6=5760 will contain duplication and if you are doing this way then to get rid of them you should divide this number by the factorial of the # of people - 4! --> \frac{5760}{4!}=240=C^2_4*2^8=favorable \ outcomes.

Consider this: there are two couples and we want to choose 2 people not married to each other. Couples: A_1, A_2 and B_1, B_2. Committees possible:

A_1,B_1; A_1,B_2; A_2,B_1; A_2,B_2.

Only 4 such committees are possible.

If we do the way you are doing we'll get: 4*2=8. And to get the right answer we should divide 8 by 2! --> 8/2!=4.

But these 4 chosen couples can send two persons (either husband or wife): 2*2*2*2

Bunuel,can you please..please explain this..im confused.... 4 chosen couples...i think we can choose 4 different people and not couples..im really confused and also how come it is 2*2*2*2...please explain

But these 4 chosen couples can send two persons (either husband or wife): 2*2*2*2

Bunuel,can you please..please explain this..im confused.... 4 chosen couples...i think we can choose 4 different people and not couples..im really confused and also how come it is 2*2*2*2...please explain

We have 6 couples: A (a_1, a_2); B (b_1, b_2); C (c_1, c_2); D (d_1, d_2); E (e_1, e_2); F (f_1, f_2);

We should choose 4 people so that none of them will be married to each other.

The above means that 4 chosen people will be from 4 different couples, for example from A, B, C, D or from A, D, E, F...

The # of ways to choose from which 4 couples these 4 people will be is C^4_6=15;

Let's consider one particular group of 4 couples: {A, B, C, D}. Now, from couple A in the group could be either a_1 or a_2, from couple B in the group could be either b_1 or b_2, from couple C in the group could be either c_1 or c_2, and from couple D in the group could be either d_1 or d_2. So each couple has two options (each couple can be represented in the group of 4 people by x_1 or x_2), so one particular group of 4 couples {A, B, C, D} can give us 2*2*2*2=2^4 groups of 4 people from different couples.

One particular group of 4 couples {A, B, C, D} gives 2^4 groups of 4 people from different couples --> 15 groups give 15*2^4 groups of 4 people from different couples (total # of ways to choose 4 people so that no two will be from the same couple) .

Re: If 4 people are selected from a group of 6 married couples [#permalink]
19 Jan 2012, 00:13

Total possible selection = 12!/(4!*8!)= 11*45 (after simplification) Favourble out come can be obtained by the multiplying the following combinations. 1. We require only 4 people. So these 4 are going to be from 4 different groups. Total availbale grops =6. So this combination is 6c4 = 6!/(4!*2!) =15 2. Select 1 member from each group = 2c1*2c1*2c1*2c1=2^4=16

Re: If 4 people are selected from a group of 6 married couples [#permalink]
27 Dec 2012, 18:34

bibha 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? 1/33 2/33 1/3 16/33 11/12

If we are to select 4 people from 6 couples WITHOUT any restriction, how many ways can we make the selection? 12!/4!6! = 11*5*9 = 495 If we are to select 4 people from 6 couples WITH restriction that no married couple can both make it to the group, only a representative? 6!/4!2! = 15 But we know that to select a person from each couple, take 2 possibilities 15*2*2*2*2 = 240

Probability = Desired/All Possibilities = 240/495 = 16/33

Re: If 4 people are selected from a group of 6 married couples [#permalink]
06 Feb 2014, 06:06

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

I believe that I had seen elsewhere that IF we were doing this same problem without the probability part of the question, we would have to divide (12x10x8x6) with 4!. Why is that not applicable when doing probability? Don't we still need the favorable outcomes?

Re: If 4 people are selected from a group of 6 married couples [#permalink]
23 Apr 2014, 22:49

Also Bunuel...One of my pain points in PnC is when the number of things to be allocated is more than the number of people...Say I have 100 pencils to be distributed 10 students..Such that each can get anything between 1 to 100.What is the formulaic approach to such questions? _________________

Appreciate the efforts...KUDOS for all

Last edited by JusTLucK04 on 24 Apr 2014, 22:34, edited 1 time in total.

Re: If 4 people are selected from a group of 6 married couples [#permalink]
24 Apr 2014, 18:20

JusTLucK04 wrote:

Also Bunuel...One of my pain points in PnC is when the number of things to be allocated is more than the number of people...Say I have 100 pencils to be distributed 10 students..Such that each can get anything between 1 to 100.What is the formulaic approach to such questions?

I would let Bunuel answer this but my thought would be: treat it as equal distribution and 100c10?

Re: If 4 people are selected from a group of 6 married couples [#permalink]
24 Apr 2014, 23:02

russ9 wrote:

JusTLucK04 wrote:

Also Bunuel...One of my pain points in PnC is when the number of things to be allocated is more than the number of people...Say I have 100 pencils to be distributed 10 students..Such that each can get anything between 1 to 100.What is the formulaic approach to such questions?

I would let Bunuel answer this but my thought would be: treat it as equal distribution and 100c10?

I think it should be...100*99....91*90 And if the question mentions that it is possible that a student recieves not even a single pencil..I think we go case wise with 1 student gets all..2 student get all pencils..and so on _________________

Re: If 4 people are selected from a group of 6 married couples [#permalink]
26 Apr 2014, 01:38

1. Select the 1st person: sure there is 1st'spouse in the group --> then 11 left 2. Select the 2nd person: probability not choose 2rd's spouse is 10/11 --> then 10 left (2 ppl are 1st and 2nd's spouses) 3. Select the 3rd person: probability 8/10 --> then 9 left (3ppl are 1st, 2nd and 3rd spouses) 4. Select the 4th: probability 6/9 --> Probability when choose 4 ppl = 10/11*8/10*6/9 = 11/33