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# There are 4 married couples in the ball room, 4 people are

Author Message
Intern
Joined: 15 Apr 2006
Posts: 16
Location: london

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07 May 2006, 14:05
prude_sb wrote:
Yes 8/35

We need to select one person out of each couple (2C1) of the 4 couples

Total number of ways: 2^4

Total number of ways of selecting 4 ppl out of 8 ppl is 8C4

Probability = (2*2*2*2)*(4*3*2*1)/(8*7*6*5) = 8/35

Ok you are choosing 4 people one person from each of the four couples,but this is not correct because you can choose a married couple but only that they shouldnt be the dancing pair
Example you can choose
M1W1 W3 M4
but though you have chosen M1 W1
M1 dances with W3 and M4 dances with W1 so none of them is dancing with their own spouse
Manager
Joined: 20 Mar 2005
Posts: 201
Location: Colombia, South America

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07 May 2006, 15:00
sorry no OA, I just made it up, but I think proffesor's answer is the correct one
Senior Manager
Joined: 12 Mar 2006
Posts: 365
Schools: Kellogg School of Management

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09 May 2006, 20:08
fighter wrote:
prude_sb wrote:
Yes 8/35

We need to select one person out of each couple (2C1) of the 4 couples

Total number of ways: 2^4

Total number of ways of selecting 4 ppl out of 8 ppl is 8C4

Probability = (2*2*2*2)*(4*3*2*1)/(8*7*6*5) = 8/35

Ok you are choosing 4 people one person from each of the four couples,but this is not correct because you can choose a married couple but only that they shouldnt be the dancing pair
Example you can choose
M1W1 W3 M4
but though you have chosen M1 W1
M1 dances with W3 and M4 dances with W1 so none of them is dancing with their own spouse

Yes you are right . Let me try again

Number of ways by which we can select 2 couples from 4 couples = 4C2

So the probability that the couples never dance together is
1 - 4C2/8C4 = 1 - 2*3/(2*7*5) = 66/70
VP
Joined: 29 Dec 2005
Posts: 1341

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09 May 2006, 20:16
conocieur wrote:
sorry no OA, I just made it up, but I think proffesor's answer is the correct one

after all i am your buddy
Senior Manager
Joined: 29 Jun 2005
Posts: 403

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11 May 2006, 00:24
Sorry, pushed submit twice...
Can't delete this post

Last edited by Dilshod on 11 May 2006, 00:44, edited 1 time in total.
Senior Manager
Joined: 29 Jun 2005
Posts: 403

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11 May 2006, 00:29
Hi,
ab cd ef gh suppose they are 4 married couples.
There are only 6 choices that assure that none of the chosen people are married: they are aceg, bdfh, bdeg, adeg, adfg, acfh
If a is chosen, then the second person should be c, e or h. The probability is 3/7.
If c is chosen, then the third person should be e or h. The probability is 2/6
If e is chosen, then the fourth person should be h. The probability is 1/5
Product of three probabilities = 2/7*1/3*1/5=2/105.
The same works for remaining combinations, so just multiply by 6 and get 12/105.
11 May 2006, 00:29

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