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6 married couples are present at a party. if 4 people are

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New post 06 Aug 2006, 15:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

6 married couples are present at a party. if 4 people are selected out of these 12, what is the probability that none of these people will be married to each other?

A)1/33

B) 2/33

C) 1/3

D) 16/33

E) 11/12

Please explain your answer.
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New post 06 Aug 2006, 21:09
I'm getting 28/33. But not in the answer choices.

Total possible combinations of 4 out of 12 is 495

1)combinations were we have two couples (example: AA,BB) --> 15
2)combinations were we have only one couple (example: AA,BC) --> 60

=420/495=28/33
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New post 06 Aug 2006, 21:21
IMO D

[(12*10*8*6)/4!]/12C4 = 16/33
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Last edited by freetheking on 06 Aug 2006, 21:27, edited 1 time in total.
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New post 06 Aug 2006, 21:23
The way i approached it is following:

We have total of 12 people.

M1 W1 M2 W2 ...... M6 W6

number of ways to choose 4 people out of 12 is 12C4 = 495

Let's say first place goes to M1.

second place can go to anybody except for W1. number of ways = 10.

third place can go to any of the remaining 8 people. number of ways = 8.

fourth place can go to any of the six eligible people. number of ways = 6.

I am not sure whether this is the right approach. If yes, how should i proceed further?
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New post 06 Aug 2006, 21:28
shoonya wrote:
The way i approached it is following:

We have total of 12 people.

M1 W1 M2 W2 ...... M6 W6

number of ways to choose 4 people out of 12 is 12C4 = 495

Let's say first place goes to M1.

second place can go to anybody except for W1. number of ways = 10.

third place can go to any of the remaining 8 people. number of ways = 8.

fourth place can go to any of the six eligible people. number of ways = 6.

I am not sure whether this is the right approach. If yes, how should i proceed further?

First place you can choose anyone.. = 12
12*10*8*6
Then remove redundancies.
For ex.
M1M2M3M4 = M2M3M4M1 = M4M3M2M1 ....
W1M2W3M4 = M2M4W1W3.....

12*10*8*6 / 4!

I believe there's easier way to solve this.
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New post 06 Aug 2006, 21:28
king, why did you divide by 4!?
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New post 06 Aug 2006, 21:30
king, please neglect my previous query. I think we both posted at almost the same time. So, your new post didn't update on my screen.

thanks for taking time to explain. This is my weakest part in maths.
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New post 06 Aug 2006, 22:13
D


number of ways to choose 4 people out of 12 is 12C4 = 495

Four people can be selected as ( keeping in mind no one is married to each other) :

4 men OR 4 women OR (1 man and 3women) OR ( 2 man + 2 woman) OR ( 3 men and 1 woman)

In case of probability
OR is same as +
and is same as *

So,
6C4 + 6C4 + (6C1 * 5C3) + ( 6C2 * 4C2 ) + ( 6C3 * 3C1 ) = 240


probability = 240/495 = 16/33
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New post 07 Aug 2006, 04:59
thanks for the detailed explanation AK.

The OA is D.

Could you tell me what's wrong with my approach?
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New post 07 Aug 2006, 13:28
Consider
AB CD EF GH IJ KL

4 can be selected out of 12 in 12C4=12*11*10*9/1*2*3*4=55*9=495 ways

If A is selected as one of 4
Then 3 can be selected from remaining 11 in 11C3 =11*10*9/1*2*3=11*5*3=165 ways of which likelyhood of B being in selection is =165/11 = 15
Therefore 150/165= 10/11 is the probability that if A is first, B is not in the list of 4.
not sure how to proceed...
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New post 07 Aug 2006, 13:38
# of ways of selecting 4 out of 12 = 12x11x10x9/4x3x2

# of ways of selecting non-couple 4 people out of 12 =12x10x8x6/4x3x2

Solving Prob = 16/33

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New post 07 Aug 2006, 13:51
Number of ways of choosing 4 people from 12= 12C4=99*5

Number of ways of choosing 4 people, none of whom are married to any other:

Choose 4 couples from 6-------------- 6C4=15
Choose man or wife from each couple ---- 2^4


So probability is 2^4*15/99*5= 48/99=16/33
  [#permalink] 07 Aug 2006, 13:51
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