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

Question

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

Bunuel · Accepted Answer

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} Answer: D. 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.

metallicafan · Answer

+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}

bibha · Answer

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
????

Bunuel · Answer

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.

ssgomz · Answer

awesome explanation +1

harithakishore · Answer

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

Bunuel · Answer

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.

harithakishore · Answer

Thats a fantabulous explanation...thankyou so much....

lansac · Answer

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

mbaiseasy · Answer

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

Answer: D

bankerboy30 · Answer

Bunuel - what's wrong with the following approach:

6c2 - two couples 15
6c1 * 10c2 - 270

1-(270/495+15/495) = 210/495= 14/33??

Posted from my mobile device

Bunuel · Answer

10C2 can also give second couple which is already counted from 6C2.

joondez · Answer

What is the chance that four chosen have nobody married to each other? Well we want to find  \frac{chance of success}{number of outcomes} . Without using combitronics:

Chance of success is 12*10*8*6
Total outcomes are 12*11*9*8

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

GyMrAT · Answer

Total # of ways to select 4 people out of 12 people = 12C4 = (12*11*10*9)/1*2*3*4) = 11 * 5 * 9

Favorable outcomes = # of ways to select 4 people out of 6 couples such that none of them are married to each other

The First person can be chosen in 12 ways.
The Second person can be chosen in 10 ways, since we will remove the spouse of first person selected & then choose.
The Third person can be chosen in 8 ways, since we will remove the spouse of second person selected, as well & then choose.
The Fourth person can be chosen in 6 ways, since we will remove the spouse of third person selected, as well & then choose.

Now we need the # of ways of selection & not the arrangement of the persons, hence order does not matter. To account for this we divide by 4!, which is the # of possible arrangements of 4 persons.

Hence Favorable outcomes = (12 * 10 * 8 * 6)/4! = 240

Required Probability = 240/(11 * 5 *9) = 16/33

Answer D.

Thanks,
GyM

j.shivank · Answer

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

Can please help in pointing out why selecting 6 individuals from 6 couples first, is wrong?

chetan2u · Answer

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

sharathnair14 · Answer

There's an easy and efficient way to solve these kinds of Combination Questions.

First, You'll have to find the total possible ways to arrange 12 people into 4 places in the committee. This is without repetition, so the seats will be filled in the following ways.

12 , 11 , 10 , 9  => So, a total of  12*11*10*9  ways.

Then, you're supposed to find the number of events where these four spots are filled in such a way that among the 12 people who are divided into 6 couples, each seat will be filled only by one representative from among the couples.

12 ,10 , 8 , 6  => So,  12*10*8*6  ways.

To find the probability, 
 \frac{12*10*8*6}{12*11*10*9}

Simplify and arrive at an answer of  \frac{16}{33}

aniruddharora · Answer

Why does this not work ?

(12C4 - 6C2) / 12C4

I am subtracting the number of ways in which the chosen 4 will always be a couple

Bunuel · Answer

6C2 is not correct. For the opposite case, we can have one couple and two people who are not married to each other, or two couples. Hence, it should be:

1- \frac{6C1*5C2*2^2 + 6C2}{12C4}=\frac{16}{33} .

As you can see, for this question, it's better to use the direct approaches shown in the topic.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards