GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 07 Jul 2020, 07:12

### 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

# If 4 people are selected from a group of 6 married couples

Author Message
TAGS:

### Hide Tags

Manager
Joined: 14 Apr 2010
Posts: 124
If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

13 Aug 2010, 07:42
6
1
73
00:00

Difficulty:

55% (hard)

Question Stats:

70% (02:22) correct 30% (02:30) wrong based on 729 sessions

### HideShow timer Statistics

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. 1/33
B. 2/33
C. 1/3
D. 16/33
E. 11/12
Math Expert
Joined: 02 Sep 2009
Posts: 65061

### Show Tags

13 Aug 2010, 08:10
9
25
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

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 $$C^4_{12}$$.

$$P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}$$

Similar problems with different approaches:
combination-permutation-problem-couples-98533.html?hilit=married%20couples
ps-combinations-94068.html?hilit=married%20couples
committee-of-88772.html?hilit=married%20couples

Hope it helps.
_________________
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 870
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

19 Feb 2012, 08:48
21
5
+1 D

A faster way to solve it:

$$\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}$$
_________________
"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 65061

### Show Tags

13 Sep 2010, 20:20
7
4
harithakishore wrote:
But these 4 chosen couples can send two persons (either husband or wife): 2*2*2*2

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) .

Hope it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 65061

### Show Tags

14 Aug 2010, 07:27
6
8
bibha wrote:
Why can't we do it like this:

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.

Hope it helps.
_________________
Math Expert
Joined: 02 Aug 2009
Posts: 8752
Re: If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

09 Mar 2019, 19:00
2
j.shivank wrote:
Dear Bunuel chetan2u

What if we first select 1 individual from each set of couple and then out of those individuals, select 4 individuals.

(2C1 * 2C1 * 2C1 * 2C1 * 2C1 * 2C1) * 6C4
= 960

Then,
Total no. of ways to choose 4 people out of 12 is = 12C4
= 495

But turns out that this method is wrong as the probability comes out to be greater than 1

There are repetitions in this..
One example..
Say 1-2, 3-4,5-6,7-8,9-10,11-12 are pair..
Select one out of each 1,3,5,7,9,11 choose ways of choosing 4: two ways would be 1,3,5,7 or 1,3,5,9
Now, choose one as 1,3,5,7,9,12 4: two ways would be 1,3,5,7 or 1,3,5,9
Not only these two, there will be many repetitions..
That is why you get probability more than 1..

But if you take 12 people first, remove the partner of the one you have chosen ..
So 10 left, close one , and again remove the partner..
Now you are left with 8...
And so on
_________________
Retired Moderator
Joined: 17 Sep 2013
Posts: 315
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
Re: If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

24 Apr 2014, 23:02
1
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
Intern
Joined: 13 Oct 2014
Posts: 1
Concentration: Marketing, General Management
GPA: 3.5
WE: Sales (Pharmaceuticals and Biotech)
Re: If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

15 Dec 2014, 06:21
1
Bunnel,
Whats wrong with this approach ?
Case-1
12c4-6c2 ( Both couples selected)
Case-2
12c4- 6c1 * 10c1 * 8c1( One couple & 2 Non-couple )

Sum up the putcomes of both the cases & find probability.

Please tell how to approach this
Retired Moderator
Joined: 25 Feb 2013
Posts: 1123
Location: India
GPA: 3.82
Re: If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

16 Dec 2017, 08:41
1
Erjan_S wrote:
metallicafan wrote:
+1 D

A faster way to solve it:

$$\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}$$

Isn't it a short and great approach? is there anything wrong with it? Why for problems like this is combinatorics a preferred approach?

Hi Erjan_S

There is nothing wrong with the approach. As the numbers are simple so you can simply count the possible and total outcomes and then keep on removing the pairs. There could be multiple ways to solve counting problems. P & C is preferred because you get the value through direct formula. in case of simple numbers you can manually count the number of possible ways or selections, but when the numbers or situation become a bit complex then you will have to rely upon P & C.
Manager
Joined: 14 Apr 2010
Posts: 124

### Show Tags

14 Aug 2010, 06:53
1
Why can't we do it like this:

total ways of selective 4 ppl from 6 married couples = 12C4
Favorable outcome = 12 *10*8*6
????
Intern
Joined: 29 Dec 2009
Posts: 18

### Show Tags

16 Aug 2010, 11:36
1
awesome explanation +1
Intern
Joined: 04 Aug 2010
Posts: 20

### Show Tags

13 Sep 2010, 19:38
But these 4 chosen couples can send two persons (either husband or wife): 2*2*2*2

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
Intern
Joined: 04 Aug 2010
Posts: 20

### Show Tags

13 Sep 2010, 20:27
Thats a fantabulous explanation...thankyou so much....
Intern
Joined: 19 Dec 2011
Posts: 4
Re: If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

19 Jan 2012, 00:13
1
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

Probability = (15*16)/(11*45)=16/33
Senior Manager
Joined: 13 Aug 2012
Posts: 386
Concentration: Marketing, Finance
GPA: 3.23
Re: If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

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

Manager
Joined: 15 Aug 2013
Posts: 223
Re: If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

23 Apr 2014, 19:42
metallicafan wrote:
+1 D

A faster way to solve it:

$$\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}$$

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?
Retired Moderator
Joined: 17 Sep 2013
Posts: 315
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
Re: If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

23 Apr 2014, 22:28
Number of ways to select...Atleast one married couple is

6C1- Choose 1 married couple for 2 seats
10C2- Chose 2 members from the remaining 5 couples for the remaining 2 seats

6C1*10C2= 270
No of ways to select 4 people from 12=12C4

P= 270/495

Answer=1- the prob of the above
=5/11

What am I missing here?
Retired Moderator
Joined: 17 Sep 2013
Posts: 315
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
Re: If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

Updated on: 24 Apr 2014, 22:34
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?

Originally posted by JusTLucK04 on 23 Apr 2014, 22:49.
Last edited by JusTLucK04 on 24 Apr 2014, 22:34, edited 1 time in total.
Manager
Joined: 15 Aug 2013
Posts: 223
Re: If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

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?
Intern
Joined: 24 Apr 2014
Posts: 8
Re: If 4 people are selected from a group of 6 married couples  [#permalink]

### Show Tags

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

Go to page    1   2    Next  [ 30 posts ]