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Committees - Slightly different three questions

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Committees - Slightly different three questions [#permalink] New post 12 Aug 2009, 13:30
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1. If a committee of 4 people is to be selected from among 6 married couples so that the
committee does not include two people who are married to each other, how many such
committees are possible? (None of them are married to each other)

2. If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes exactly two people who are married to each other, how many such
committees are possible? (One couple and other two non couple)

3. If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes four people who are married to each other, how many such
committees are possible? (Both couple)

I know the answer of first question. Don't know the OA for 2 and 3.
Please elaborate your answers. Any specific approach to these kinds of questions will be appreciated.
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Re: Committees - Slightly different three questions [#permalink] New post 12 Aug 2009, 20:40
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1.

a) Choose 4 couples for committee: C_4^6

b) Choose one person for each couple: (C_1^2)^4

c) The number of possible committee: C_4^6 * (C_1^2)^4 = \frac{6*5}{2} * 2^4 = 15 * 16 = 240

2.

a) Choose one couple for committee: C_1^6

b) Choose 2 couples for committee from which we are going to choose 2 people: C_2^5

c) Choose one person for each couple out of 2 couples: (C_1^2)^2

d) The number of possible committee: C_1^6*C_2^5 * (C_1^2)^2 = 6*\frac{5*4}{2} * 2^2 = 240

3.

a) Choose 2 couples for committee: C_2^6 = \frac{6*5}{2} = 15
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Re: Committees - Slightly different three questions [#permalink] New post 12 Aug 2009, 21:12
walker wrote:
1.

a) Choose 4 couples for committee: C_4^6

b) Choose one person for each couple: (C_1^2)^4

c) The number of possible committee: C_4^6 * (C_1^2)^4 = \frac{6*5}{2} * 2^4 = 15 * 16 = 240

2.

a) Choose one couple for committee: C_1^6

b) Choose 2 couples for committee from which we are going to choose 2 people: C_2^5

c) Choose one person for each couple out of 2 couples: (C_1^2)^2

d) The number of possible committee: C_1^6*C_2^5 * (C_1^2)^2 = 6*\frac{5*4}{2} * 2^2 = 240

3.

a) Choose 2 couples for committee: C_2^6 = \frac{6*5}{2} = 15


Thanks for explaining!
Could anyone tell me where I can find practice questions on Number Theory and Probability on the forum?

Thanks in advance
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Re: Committees - Slightly different three questions [#permalink] New post 12 Aug 2009, 21:59
walker, you rock. You are a master of such questions!
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Re: Committees - Slightly different three questions [#permalink] New post 12 Aug 2009, 22:34
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Could anyone tell me where I can find practice questions on Number Theory and Probability on the forum?

Thanks in advance


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Ibodullo wrote:
walker, you rock. You are a master of such questions!


Thanks! :) It was one of my weaknesses when I began with GMAT...
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Re: Committees - Slightly different three questions [#permalink] New post 13 Aug 2009, 06:44
walker wrote:
gauravc wrote:
Could anyone tell me where I can find practice questions on Number Theory and Probability on the forum?

Thanks in advance


See link in my signature

Ibodullo wrote:
walker, you rock. You are a master of such questions!


Thanks! :) It was one of my weaknesses when I began with GMAT...


Walker, great to see you replying to these posts. You are simply too good for GMAT Quant. I wonder how you did not get 51.
Thanks for the excellent explanation.
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Re: Committees - Slightly different three questions [#permalink] New post 13 Aug 2009, 08:05
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... I wonder how you did not get 51.


There were two hows:
- a few silly mistakes. I even still remember one. Very silly I even thought that GMAC made a mistake :) but in a few hours I realized that it was my fall.
- as a non-native English speaker I went through quant section waiting for a real battle - verbal. So, I kept my energy for it.
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Re: Committees - Slightly different three questions [#permalink] New post 13 Aug 2009, 16:27
walker wrote:
getmba wrote:
... I wonder how you did not get 51.


There were two hows:
- a few silly mistakes. I even still remember one. Very silly I even thought that GMAC made a mistake :) but in a few hours I realized that it was my fall.
- as a non-native English speaker I went through quant section waiting for a real battle - verbal. So, I kept my energy for it.


I would say your second "how - saving energy for verbal" paid off handsomely (V40). Nonetheless you deserved it.
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Re: Committees - Slightly different three questions [#permalink] New post 14 Aug 2009, 03:48
Quote:
getmba:
1. If a committee of 4 people is to be selected from among 6 married couples so that the
committee does not include two people who are married to each other, how many such
committees are possible? (None of them are married to each other)


Quote:
Walker:
1.
a) Choose 4 couples for committee:


I am confused here. Is it 4 couple or 4 people. I calculated for 4 people and I got 425. I know i am wrong somewhere. Walker, would you be kind enough to write a small note explaining it if it were infact 4 people, how the solution can be found. Thanks for the help. :)
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Re: Committees - Slightly different three questions [#permalink] New post 14 Aug 2009, 05:16
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bhanushalinikhil wrote:
I am confused here. Is it 4 couple or 4 people. I calculated for 4 people and I got 425. I know i am wrong somewhere. Walker, would you be kind enough to write a small note explaining it if it were infact 4 people, how the solution can be found. Thanks for the help. :)


The key to combination problems is to understand way how we can construct our resulting set. Look at my steps as "construction" stages:

1) choose 4 couple out of 6
2) choose 1 person out of 2 for each couple.

Actually, I construct set with 4 people (None of them are married to each other).

At the same time, we can choose another "construction" scheme:

1) choose 4 people out of 6*2 people step by step with restriction on marriage.
1.1. First person: no restriction ---> 12 possibilities
1.2. Second person: he or she cannot be spouse of first person ---> 10 possibilities
1.3. Third person: he or she cannot be spouse of first or second people ---> 8 possibilities
1.4. Fourth person: he or she cannot be spouse of first, second or third people ---> 6 possibilities

N = 12 * 10 *8 * 6

2) Now, we must exclude possible permutation in the committee because: ABCD is the same committee as BACD --> P_4^4 = 4!

So, \frac{12 * 10 *8 * 6}{4*3*2} = 240

As you can see, using different "construction" approach led us to the same results
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Re: Committees - Slightly different three questions [#permalink] New post 14 Aug 2009, 10:29
walker wrote:
gauravc wrote:
Could anyone tell me where I can find practice questions on Number Theory and Probability on the forum?

Thanks in advance


See link in my signature


Thanks a ton :!:
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Re: Committees - Slightly different three questions [#permalink] New post 08 Jun 2010, 20:36
Quote:
2. If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes exactly two people who are married to each other, how many such
committees are possible? (One couple and other two non couple)

3. If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes four people who are married to each other, how many such
committees are possible? (Both couple)

When I used the same construction mentioned above by Walker, I got different results:
2. Choose 1st person = 12 possibilities
choose his spouse = 1 possibility
choose 3rd person =10 possibilities
choose 4th person = 8 possibilities
C= 12*1*10*8/4!= 40

Is it possible to choose 3 persons not married to each other, then choose one of the three spouses?

3. Choose 1st person = 12 possibilities
choose his spouse = 1 possibility
choose 3rd person =10 possibilities
choose his spouse = 1 possibility
C= 12 *10/4!=5

Obviously, there are flaws in my logic but I can't see it.Please help.
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Re: Committees - Slightly different three questions [#permalink] New post 12 Sep 2012, 05:06
getmba wrote:

1. If a committee of 4 people is to be selected from among 6 married couples so that the
committee does not include two people who are married to each other, how many such
committees are possible? (None of them are married to each other)


ways of choosing one couple = 1*1* 10C2 =45 since there are six couples , so 45*6 = 270 ways
total ways of choosing 4 people out of 12 = 12c4= 495

hence no couple together = 495-270= 225

answer not matching with any given in this topic.what am I doing wrong if any ?

getmba wrote:
2. If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes exactly two people who are married to each other, how many such
committees are possible? (One couple and other two non couple)



ways of choosing one couple = 1*1* 10C2 =45 since there are six couples , so 45*6 = 270 ways


getmba wrote:
3. If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes four people who are married to each other, how many such
committees are possible? (Both couple)


6C2= 15
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Re: Committees - Slightly different three questions [#permalink] New post 12 Sep 2012, 12:07
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stne wrote:
getmba wrote:

1. If a committee of 4 people is to be selected from among 6 married couples so that the
committee does not include two people who are married to each other, how many such
committees are possible? (None of them are married to each other)


ways of choosing one couple = 1*1* 10C2 =45 since there are six couples , so 45*6 = 270 ways
total ways of choosing 4 people out of 12 = 12c4= 495

hence no couple together = 495-270= 225

answer not matching with any given in this topic.what am I doing wrong if any ?

getmba wrote:
2. If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes exactly two people who are married to each other, how many such
committees are possible? (One couple and other two non couple)



ways of choosing one couple = 1*1* 10C2 =45 since there are six couples , so 45*6 = 270 ways


getmba wrote:
3. If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes four people who are married to each other, how many such
committees are possible? (Both couple)


6C2= 15


Q3: Your answer is correct as in fact two couples are chosen as the 4 members of the committee.

Q1 and Q2: For one couple and the other two people not a couple "ways of choosing one couple = 1*1* 10C2 =45" is not correct.
10C2 means choose 2 people out of 10. Any 10 people include at least 4 couples (you can choose one person from each couple, which is a maximum of 6 different people, but for the next 4 you have no choice but starting to choose spouses for some of the previous ones).
Where those 1's came from?
If for Q1 you meant to subtract from the total number of possibilities of choosing any 4 people for the committee the number of possibilities when there is at least one couple chosen, you should have considered the cases when there is exactly one couple and two not married to each other and the case when there are exactly two married couples.

From the posted solutions, you can see that number of possibilities when no (0 zero) couple is chosen is 240, when exactly one couple is chosen again 240 possibilities, and two couples chosen 15, which gives a total of 495 = 12C4.

In the above questions all the time you have to choose first couple(s) instead of choosing persons. When you need a couple, you just choose a couple. When you need people without their spouses, you again choose the couples from which then you can choose either spouse.
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Re: Committees - Slightly different three questions [#permalink] New post 12 Sep 2012, 20:28
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stne wrote:

1. If a committee of 4 people is to be selected from among 6 married couples so that the
committee does not include two people who are married to each other, how many such
committees are possible? (None of them are married to each other)

ways of choosing one couple = 1*1* 10C2 =45 since there are six couples , so 45*6 = 270 ways
total ways of choosing 4 people out of 12 = 12c4= 495

hence no couple together = 495-270= 225

answer not matching with any given in this topic.what am I doing wrong if any ?



Here is the error:

ways of choosing one couple = 1*1* 10C2 =45
You have taken the first couple (1*1) and then selected 2 people out of the remaining 10. But, mind you, when you do 10C2, you are including the number of ways in which you get another couple too.

Say the couples are:
A1, A2
B1, B2
C1, C2
D1, D2
E1, E2
F1, F2
You select A1, A2 and then you select 2 of the remaining 10. You will get (B1, C2), (C1, D1) etc but you will also get (B1, B2), (C1, C2) etc as the remaining two.
These 45 selections you get include those where you have both the couples. This is not a problem since what you actually need to find here is the number of ways in which there is AT LEAST one couple (the complement of no couple).
But the problem arises when you multiply this by 6 since when you select (A1, A2) and get (B1, B2) as the remaining two, you get the same 4 people when you select (B1, B2) and get (A1, A2) as the remaining two.
Hence there is double counting in your 270 ways.

When you select (B1, B2), you cannot get (A1, A2) anymore since you already accounted for it when you selected (A1, A2).
When you select (C1, C2), you cannot get (A1, A2) and (B1, B2) anymore since you already accounted for them when you selected (A1, A2) and later (B1, B2)
etc
You need to remove 1+2+3+4+5 = 15 cases to avoid double counting.

No of ways in which there is AT LEAST one couple = 270 - 15 = 255

Now, 12C2 = 495
No of ways in which there are no couples is 495 - 255 = 240 (answer)

You should instead try the more straight forward way:
Select 4 of the 6 couples in 6C4 = 15 ways.
Take one person from each couple in 2*2*2*2 ways.
Total no of ways = 15*16 = 240

getmba wrote:
2. If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes exactly two people who are married to each other, how many such
committees are possible? (One couple and other two non couple)


ways of choosing one couple = 1*1* 10C2 =45 since there are six couples , so 45*6 = 270 ways


Here, your 45 includes 5 cases in which you have both couples. You select (A1, A2) and get (B1, B2) or (C1, C2) etc as the remaining two.
No of ways of getting ONLY ONE couple = 45 - 5 = 40
Since there are 6 couples, you get 40*6 = 240 (answer)
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Re: Committees - Slightly different three questions [#permalink] New post 13 Sep 2012, 04:44
VeritasPrepKarishma wrote:
stne wrote:

1. If a committee of 4 people is to be selected from among 6 married couples so that the
committee does not include two people who are married to each other, how many such
committees are possible? (None of them are married to each other)

ways of choosing one couple = 1*1* 10C2 =45 since there are six couples , so 45*6 = 270 ways
total ways of choosing 4 people out of 12 = 12c4= 495

hence no couple together = 495-270= 225

answer not matching with any given in this topic.what am I doing wrong if any ?



Here is the error:

ways of choosing one couple = 1*1* 10C2 =45
You have taken the first couple (1*1) and then selected 2 people out of the remaining 10. But, mind you, when you do 10C2, you are including the number of ways in which you get another couple too.

Say the couples are:
A1, A2
B1, B2
C1, C2
D1, D2
E1, E2
F1, F2
You select A1, A2 and then you select 2 of the remaining 10. You will get (B1, C2), (C1, D1) etc but you will also get (B1, B2), (C1, C2) etc as the remaining two.
These 45 selections you get include those where you have both the couples. This is not a problem since what you actually need to find here is the number of ways in which there is AT LEAST one couple (the complement of no couple).
But the problem arises when you multiply this by 6 since when you select (A1, A2) and get (B1, B2) as the remaining two, you get the same 4 people when you select (B1, B2) and get (A1, A2) as the remaining two.
Hence there is double counting in your 270 ways.

When you select (B1, B2), you cannot get (A1, A2) anymore since you already accounted for it when you selected (A1, A2).
When you select (C1, C2), you cannot get (A1, A2) and (B1, B2) anymore since you already accounted for them when you selected (A1, A2) and later (B1, B2)
etc
You need to remove 1+2+3+4+5 = 15 cases to avoid double counting.

No of ways in which there is AT LEAST one couple = 270 - 15 = 255

Now, 12C2 = 495
No of ways in which there are no couples is 495 - 255 = 240 (answer)

You should instead try the more straight forward way:
Select 4 of the 6 couples in 6C4 = 15 ways.
Take one person from each couple in 2*2*2*2 ways.
Total no of ways = 15*16 = 240

getmba wrote:
2. If a committee of 4 people is to be selected from among 6 married couples so that the
committee includes exactly two people who are married to each other, how many such
committees are possible? (One couple and other two non couple)


ways of choosing one couple = 1*1* 10C2 =45 since there are six couples , so 45*6 = 270 ways


Here, your 45 includes 5 cases in which you have both couples. You select (A1, A2) and get (B1, B2) or (C1, C2) etc as the remaining two.
No of ways of getting ONLY ONE couple = 45 - 5 = 40
Since there are 6 couples, you get 40*6 = 240 (answer)


I have understood the 6c4 *(2)^4 = 240

Thank you Both Karishma and Eva .. Yes the 10C2 was unlogical , it does include cases where couples could be selected

I was confused with another case : where we have 6 people including Ann and Beth in how many ways can we select 3 members so that Ann and Beth are not members of the committee.

total ways to select 3 out of 6 = 6c3 = 20

now cases where Ann and Beth are together = 1*1*4C1 = 4

hence cases where they are not together = 20 - 4 = 16


In this 6 couples sum , there is no logic of assuming 1*1* 10C2 = The ones make no sense as pointed out by EVA , the ones are ok for the Ann and Beth example above , as we know one way to choose Ann and one way to choose Beth.

although for the six couple sum we can also assume Ann and Beth as one of the couple ( among the six ) and solve as Karishma had pointed out, then the 1's seem logical.( although little complex this way )


Now for the Six couple example The alternate way would be :

Total ways to select 4 people such that no one is married to each other = Total ways to select 4 people - ( Total ways to select only ONE couple + total ways to select exactly two couple )

to select exactly one couple = 12*1*10*8/2!2! = 240 ( [CC][NN] , two of same kind and two of other kind )

two select exactly two couple = 12*1*10*1/ 2!2!2! = 15 { [C1C2] [B1B2] } now the order of C1 and C2 does not matter hence dividing by 2!
similarly the order of B1 and B2 does not matter hence another 2! division

Also the order of the two couples themselves does not matter we could have [B1B2] first and then [C1C2] hence another 2! division.

hence 4 people where none are married to each other = 495 - ( 240+15) = 240

Now karishma and Eva and others I could never have arrived at 15 for selecting exactly two couples , if I didn't know the answer already.
That extra 2! for the arrangement of the 2 couples ( not the members in a couple ) would not occur to me .

is this correct thinking ? is the logic correct ?

Just for understanding sake , What is the way to calculate exactly two couple by fundamental counting method ( Not the 6c2 way ) ?
But 12 *1 *10 *1 way


Now for exactly one couple lets say we have CCNN where CC is one couple and NN are not couples here too we could argue that we can have [NN] first then [CC] hence another 2! was necessary .( we already have 2! for members of CC and 2! for members of NN but to nullify the order of 1 couple and 1 non couple shouldn't we have another 2! in the denominator for { CCNN and NNCC } ,as we did for the two couples).

So for the exactly one couple case why can't we have 12*1*10*8/ 2!2!2! or can we ?

So basically what is the logic of exactly one couple fundamental counting method .

I think if Karishma , Eva or any of you could provide just a little help here , I could get it fully.
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Re: Committees - Slightly different three questions [#permalink] New post 13 Sep 2012, 22:07
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stne wrote:
Thank you Both Karishma and Eva .. Yes the 10C2 was unlogical , it does include cases where couples could be selected

I was confused with another case : where we have 6 people including Ann and Beth in how many ways can we select 3 members so that Ann and Beth are not members of the committee.

total ways to select 3 out of 6 = 6c3 = 20

now cases where Ann and Beth are together = 1*1*4C1 = 4

hence cases where they are not together = 20 - 4 = 16


In this 6 couples sum , there is no logic of assuming 1*1* 10C2 = The ones make no sense as pointed out by EVA , the ones are ok for the Ann and Beth example above , as we know one way to choose Ann and one way to choose Beth.

although for the six couple sum we can also assume Ann and Beth as one of the couple ( among the six ) and solve as Karishma had pointed out, then the 1's seem logical.( although little complex this way )



Hope you understand that the entire approach of assuming the 12 people independently is cumbersome. The only reason I have solved it that way for you is to help you understand where you went wrong when you solved it that way. It's ok if you want to understand fully how to solve using that approach because if you want to take that approach in some other question, you should know how to do it properly.


stne wrote:
Now for the Six couple example The alternate way would be :

Total ways to select 4 people such that no one is married to each other = Total ways to select 4 people - ( Total ways to select only ONE couple + total ways to select exactly two couple )

to select exactly one couple = 12*1*10*8/2!2! = 240 ( [CC][NN] , two of same kind and two of other kind )

to select exactly two couple = 12*1*10*1/ 2!2!2! = 15 { [C1C2] [B1B2] } now the order of C1 and C2 does not matter hence dividing by 2!
similarly the order of B1 and B2 does not matter hence another 2! division

Also the order of the two couples themselves does not matter we could have [B1B2] first and then [C1C2] hence another 2! division.

hence 4 people where none are married to each other = 495 - ( 240+15) = 240

Now karishma and Eva and others I could never have arrived at 15 for selecting exactly two couples , if I didn't know the answer already.
That extra 2! for the arrangement of the 2 couples ( not the members in a couple ) would not occur to me .

is this correct thinking ? is the logic correct ?


Yes, the logic is correct though twisted.


stne wrote:
Just for understanding sake , What is the way to calculate exactly two couple by fundamental counting method ( Not the 6c2 way ) ?
But 12 *1 *10 *1 way


Using the basic counting principle, you can easily solve this. Think in terms of couples, not people.
No of ways = 6*5/2! (Select one couple in 6 ways, another in 5 ways and divide by 2 to un-arrange the couples)

stne wrote:
Now for exactly one couple lets say we have CCNN where CC is one couple and NN are not couples here too we could argue that we can have [NN] first then [CC] hence another 2! was necessary .( we already have 2! for members of CC and 2! for members of NN but to nullify the order of 1 couple and 1 non couple shouldn't we have another 2! in the denominator for { CCNN and NNCC } ,as we did for the two couples).

So for the exactly one couple case why can't we have 12*1*10*8/ 2!2!2! or can we ?

So basically what is the logic of exactly one couple fundamental counting method .

I think if Karishma , Eva or any of you could provide just a little help here , I could get it fully.


Using basic counting principle for one couple:
No of ways = 6*(10*8/2!) (Select a couple in 6 ways. You have 10 people left. Select one of them. Now you have to select 1 out of 8 people. But you arranged the last two people so divide by 2!)

In this case, 12*1*10*8/ 2!2!2! is incorrect because you have an extra 2! here. You select one person out of 12 and take the partner along. You need to divide this by 2! because you could have selected the partner first instead. Then you pick one of the other 10 and one of the other 8 and divide by 2!.
So 12*10*8/2!*2! is correct
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Re: Committees - Slightly different three questions [#permalink] New post 14 Sep 2012, 02:02
Thank you karishma , Based on your explanation I have just created a few more cases , just to test my understanding

Please could you check if I have understood correctly

lets says 6 couples and we have to select 6 members

Lets find
(1) exactly one couple
(2) exactly two couple
(3) exactly 3 couple
(4 ) No couple


for (1) 6C1 *(10*8/2!)*(6*4/2!)= 6*40*12 =2880 hope this is correct

exactly Two couple

6C2 *(8*6/2!)=15*24= 360 Hope this is correct ?

Exactly 3 couple

6C3 = 20


No couple

6c6 *(2)^6 = 64

The answer for question 1 does not look correct , although we followed the same process when we had to select 4 members

if we had to select 4 members , from 6 couples , such that we had only 1 couple ( as we did previously).
We did 6C1*(10*8/2!)= 240

but for 1 couple when we have to select 6 members ? I did 6C1 *(10*8/2!)*(6*4/2!)= 2880 (following same logic) looks incorrect.
Not sure about (2) either , but (3) and (4) looks correct.

Please check question (1), what could be the general way , if any?
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Re: Committees - Slightly different three questions [#permalink] New post 14 Sep 2012, 04:04
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stne wrote:
Thank you karishma , Based on your explanation I have just created a few more cases , just to test my understanding

Please could you check if I have understood correctly

lets says 6 couples and we have to select 6 members

Lets find
(1) exactly one couple
(2) exactly two couple
(3) exactly 3 couple
(4 ) No couple


for (1) 6C1 *(10*8/2!)*(6*4/2!)= 6*40*12 =2880 hope this is correct

exactly Two couple

6C2 *(8*6/2!)=15*24= 360 Hope this is correct ?

Exactly 3 couple

6C3 = 20


No couple

6c6 *(2)^6 = 64

The answer for question 1 does not look correct , although we followed the same process when we had to select 4 members

if we had to select 4 members , from 6 couples , such that we had only 1 couple ( as we did previously).
We did 6C1*(10*8/2!)= 240

but for 1 couple when we have to select 6 members ? I did 6C1 *(10*8/2!)*(6*4/2!)= 2880 (following same logic) looks incorrect.
Not sure about (2) either , but (3) and (4) looks correct.

Please check question (1), what could be the general way , if any?


for (1) 6C1 *(10*8/2!)*(6*4/2!)= 6*40*12 =2880 hope this is correct

No. Choose on couple out of the 6 and take both persons, this is 6C1. Then, four the next 4 single people you have 10*8*6*4 possibilities, but you have to divide by 4!, because the order in which you choose them doesn't matter.
You divide by the factorial of the number of chosen single persons, not all the time by 2!.

exactly Two couple
6C2 *(8*6/2!)=15*24= 360 Hope this is correct ?
Yes, correct.

Exactly 3 couple
6C3 = 20
Yes, correct

No couple
6c6 *(2)^6 = 64
Yes, correct
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Re: Committees - Slightly different three questions [#permalink] New post 14 Sep 2012, 07:36
Thank you Eva that cleared all doubts

was going wrong for 1 couple

now I get it
6C1 *(10*8*6*4 /4!) = 480

Highly Appreciated !
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Re: Committees - Slightly different three questions   [#permalink] 14 Sep 2012, 07:36
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