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

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If 4 people are selected from a group of 6 married couples [#permalink] New post 13 Aug 2010, 07:42
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

75% (02:11) correct 24% (01:40) wrong based on 112 sessions
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
[Reveal] Spoiler: OA
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Re: confuseddd [#permalink] New post 13 Aug 2010, 08:10
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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}

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.
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Re: confuseddd [#permalink] New post 14 Aug 2010, 06:53
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
????
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Re: confuseddd [#permalink] New post 14 Aug 2010, 07:27
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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.
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Re: confuseddd [#permalink] New post 16 Aug 2010, 11:36
awesome explanation +1
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Re: confuseddd [#permalink] New post 13 Sep 2010, 19:38
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
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Re: confuseddd [#permalink] New post 13 Sep 2010, 20:20
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harithakishore wrote:
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) .

Hope it's clear.
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Re: confuseddd [#permalink] New post 13 Sep 2010, 20:27
Thats a fantabulous explanation...thankyou so much....
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Re: If 4 people are selected from a group of 6 married couples [#permalink] New post 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

Probability = (15*16)/(11*45)=16/33
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Re: If 4 people are selected from a group of 6 married couples [#permalink] New post 19 Feb 2012, 08:48
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+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}
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Re: If 4 people are selected from a group of 6 married couples [#permalink] New post 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

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