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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

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 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]

Show Tags

19 Jan 2012, 01:13

1

This post was BOOKMARKED

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]

Show Tags

27 Dec 2012, 19:34

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

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]

Show Tags

23 Apr 2014, 23:49

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 Don't let an extra chromosome get you down..

Last edited by JusTLucK04 on 24 Apr 2014, 23:34, edited 1 time in total.

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

Show Tags

24 Apr 2014, 19:20

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]

Show Tags

25 Apr 2014, 00:02

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
_________________

Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down..

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

Show Tags

26 Apr 2014, 02:38

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

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...