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

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There are 4 married couples in the ball room, 4 people are [#permalink] New post 04 May 2006, 10:10
There are 4 married couples in the ball room, 4 people are randomly selected to dance, what is the chance that none of the dancing partners are also married?
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reply [#permalink] New post 04 May 2006, 12:04
I am getting 6/7

total no of ways to choose 4 people out of 8 is C(8,4)=70
ways in which at least one is a partner = C(choosing one couple+C(choosing 2 couples)=C(4,1)+C(4,2)=10
probability of not choosing at least one couple=1-10/70=6/7
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 [#permalink] New post 04 May 2006, 12:18
How is it possible to select 4 unwedded people from 4 married couples? All of them are married so there would not be any unmarried one.

I'm confused. :?
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 [#permalink] New post 04 May 2006, 13:36
pesquadero wrote:
How is it possible to select 4 unwedded people from 4 married couples? All of them are married so there would not be any unmarried one.

I'm confused. :?

There are 8 people in total ( 4 couples).
example Man1 wife1
M2 W2
M3 W3
M4 W4
If you choose 4 people out of these at random you could have
M1 M2 M3 M4 or W1 W2 W3 W4 or M1 M2 W3 W4 or W1 W2 M3 M4
or M1 w1 M2 W4 (here you have chosen 4 people which includes one couple)What is asked is probability of not having even a single married pair
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Re: couples [#permalink] New post 04 May 2006, 16:07
conocieur wrote:
There are 4 married couples in the ball room, 4 people are randomly selected to dance, what is the chance that none of the dancing partners are also married?



select 1 person - his/her partner can be selected in 6 ways

select second person his/partner can be selected in 5 ways if his/her spouse is with the first person

or else 4

so 6x5 + 6x4 =30+24 = 54

54/8c4 = 54/70? oa?
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 [#permalink] New post 04 May 2006, 16:28
Four couples and 4 people have to be selected without selecting their spouse

8C1*6C1 + 7C1*5C1 +6C1*4C1 +5C1*3C1 -----(i)

Lets take this scenario
8C1*6C1

There are 8 people initially So select 1 from 8 in 8C1 ways.
Select a partner for this person in 6C1 ways 'cos the spouse shouldnt be selected.

Like wise select for all the 4 people.

Now again we have to mulitply (i) by 4 'cos
in this selection 8C1*6C1 it can be either the husband or wife so 2! ways for each selection .

So 8C1*6C1 has to be mutiplied by 4

So the probablity = 61/105
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 [#permalink] New post 04 May 2006, 17:22
kook44 wrote:
2^4 / 4C8 = 4/15?


i guess this the correct rule but 2^4/8c4 = 16/70=8/35
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 [#permalink] New post 05 May 2006, 01:24
I also get 6/7 i.e. 1- atleast one couple is married. Can you PLEASE post the OA??
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reply [#permalink] New post 05 May 2006, 02:49
on rethinking
The total number of ways you can pick 4 poeple out of 8 is 8C4=70
no of ways to choose (only)one couple and two others =4C1.6C1.4C1
Explanation :
4C1 because you are choosing one couple out of four
6 C1 for choosing the 3rd person ie one person from remaining 6
4C1 for choosing the 4 th person from .you are choosing from 4 people and not five because you eliminate the spouse of 3rd person
So the number of ways you can choose 4 people so that you have one couple and two others =4x6x4
Once we have chosen 4 people we can pair them so as to have two dancing pairs total number of ways we can pair =6 ways
and in only one of these six cases we have the married couple pairing as dancing couple
So out of 4x6x4 we have 16 cases when the married couple pairs as dancing partner
ie 4x6x4x1/6=16

We now consider cases where we choose two couples
so we get 4C2
but here we get a couple as dancing partners in two ways out of 6
So no of ways we get two couples as dancing partners=4C2x2/6=2

probability of not having at least one couple as dancing partner=
1-(number of ways of one couple as dancing partner+2couples dancing)/total
1-(16+2)/70=26/35


I know it is long ,but this is how i thought
Icould be wrong becoz I am doing my maths after 15 years !

Prof could you please explain how 2^4/8C4.
I am unable to understand how you and kook44 got that
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 [#permalink] New post 05 May 2006, 11:57
Agree with prof.
2^4/8c4 = 16/70=8/35
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Couples [#permalink] New post 05 May 2006, 20:29
There are total 4 couples i.e, 8 persons. Therefore No. of ways of selecting 4 persons from 8 persons = 8C4 = 70 ways (this includes married couples)
Now from this we need to subtract the number of ways in which we had selected the the married couples.
Consider we bunch M1W1, M2W2, M3W3, M4W4 there are 4 couples when one is selected his/her spouse is automatcally selected.
Hence number of ways of selecting a bunch of 2 couples (4 persons) from a set of 4 couples (8 persons) = 4C2 = 6 ways
Again, number of ways of selecting 1 couple and two unmarried persons
Number of ways of selecting 1 couple (2 persons) out of 4 couples= 4C1=4 ways
In addition, the other two persons are to be necessarily unmarried.
So Out of the remaining 6 persons, 1 person can be selected in 6C1 = 6 ways
and for every this person selected the person can only be selected from the balance lot i.e, 4 persons (excluding his/her spouse) in 4C1 = 4 ways.
so the total number of ways of selecting 1 couple team = 4+(6*4) = 28
Hence the total number of ways of selecting married couples into the team = 6+28 = 34
Hence the total number of ways of not selecting a married couple = 70-34 = 36
Hence chance of not selecting a couple = 36/70 = 18/35
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 [#permalink] New post 05 May 2006, 21:24
I am getting (8c1)(6c1)(4c1)(2c1)/(8c4)...what is the OA?
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 [#permalink] New post 06 May 2006, 03:57
Please ignore the earlier posting regarding the solution of the problem. it has certain mistakes. Consider this.
There are 4 positions to select from total 8 persons. That can be done in
8C1*7C1*6C1*5C1
Again, if the positions that need to be selected, should not contain any married couples, then the number of ways are 8C1*6C1*4C1*2C1.
Therefore, the chance = (8C1*6C1*4C1*2C1) / (8C1*7C1*6C1*5C1) = 8/35.
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 [#permalink] New post 06 May 2006, 09:40
shevy wrote:
I am getting (8c1)(6c1)(4c1)(2c1)/(8c4)...what is the OA?


Yes the total number of ways to select 4 people out of 8 is

8*7*6*5

so it is

(8c1)(6c1)(4c1)(2c1)/(8*7*6*5)
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 [#permalink] New post 06 May 2006, 22:09
buddy, its OA time..
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 [#permalink] New post 06 May 2006, 22:50
Probability of dancing partner are choose and dancing togather is

Probability of selecting Male*Probability of selecting female * Probability both dancing to gather

4/8 * 3/7 * 1/3 = 1/14

4/8 => Male can be picked 4 ways
3/7 => female can be picked in one of another three slote
1/3 => female picked she will dance with male

so our required probability is 1- 1/14 = 13/14
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 [#permalink] New post 07 May 2006, 01:11
kook44 wrote:
2^4 / 4C8 = 4/15?


How did you get 2^4?
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 [#permalink] New post 07 May 2006, 02:38
I am getting 3/4!



1 - 4C2/8C4 = 1 - 6/24 = 18/24 = 3/4!


Get us out of this plight!! Givge us the OA!
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 [#permalink] New post 07 May 2006, 10:18
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
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 [#permalink] New post 07 May 2006, 11:32
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


Now I understand. Thank you!
  [#permalink] 07 May 2006, 11:32
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