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Manager
Joined: 20 Mar 2005
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There are 4 married couples in the ball room, 4 people are [#permalink]
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04 May 2006, 11:10
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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?



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

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=110/70=6/7



Manager
Joined: 29 Apr 2006
Posts: 85

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.



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

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



Senior Manager
Joined: 09 Aug 2005
Posts: 283

Re: couples [#permalink]
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04 May 2006, 17:07
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?



VP
Joined: 28 Mar 2006
Posts: 1369

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



VP
Joined: 29 Dec 2005
Posts: 1340

kook44 wrote: 2^4 / 4C8 = 4/15?
i guess this the correct rule but 2^4/8c4 = 16/70=8/35



Manager
Joined: 23 Mar 2006
Posts: 75

I also get 6/7 i.e. 1 atleast one couple is married. Can you PLEASE post the OA??



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

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



Manager
Joined: 27 Jan 2006
Posts: 156
Location: Europe

Agree with prof.
2^4/8c4 = 16/70=8/35



Intern
Joined: 14 Jan 2006
Posts: 20
Location: Visakhapatnam, India

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 = 7034 = 36
Hence chance of not selecting a couple = 36/70 = 18/35



Manager
Joined: 27 Mar 2006
Posts: 136

I am getting (8c1)(6c1)(4c1)(2c1)/(8c4)...what is the OA?



Intern
Joined: 14 Jan 2006
Posts: 20
Location: Visakhapatnam, India

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.



Manager
Joined: 27 Mar 2006
Posts: 136

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)



VP
Joined: 29 Dec 2005
Posts: 1340

buddy, its OA time..



Senior Manager
Joined: 05 Jan 2006
Posts: 381

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



Manager
Joined: 29 Apr 2006
Posts: 85

kook44 wrote: 2^4 / 4C8 = 4/15?
How did you get 2^4?



VP
Joined: 29 Apr 2003
Posts: 1403

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!



Senior Manager
Joined: 12 Mar 2006
Posts: 365
Schools: Kellogg School of Management

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



Manager
Joined: 29 Apr 2006
Posts: 85

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!







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