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

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.

I would say your second "how - saving energy for verbal" paid off handsomely (V40). Nonetheless you deserved it.

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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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.

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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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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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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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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