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# A committe of three people is to be chosen from four married

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Senior Manager
Joined: 06 Apr 2008
Posts: 435
A committe of three people is to be chosen from four married [#permalink]

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26 Jul 2008, 02:36
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A committe of three people is to be chosen from four married couples . WHat is number of different committees that can be chosen if two people who are married to each other cannot both serve on committee

1. 16
2. 24
3. 26
4. 30
5. 32

Source -> GMAT prep
Manager
Joined: 24 Apr 2008
Posts: 161

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26 Jul 2008, 02:54
Ans:5

8x6x4/3! = 32
Senior Manager
Joined: 06 Apr 2008
Posts: 435

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26 Jul 2008, 03:03
iamcartic wrote:
Ans:5

8x6x4/3! = 32

Can you provide explanation please ... Thanks
Manager
Joined: 06 Feb 2008
Posts: 86

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26 Jul 2008, 04:40
let me try...out of 4 couples we have to choose three at a time....no of combinations is given by 4C3 morever within each choosen couple we have a option of choosing one out of two people or 2C1

So, 4C3*2C1*2C1*2C1 = 4*2*2*2 = 4*8 = 32

Hope this helps..!!!
Manager
Joined: 03 Jun 2008
Posts: 137
Schools: ISB, Tuck, Michigan (Ross), Darden, MBS

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27 Jul 2008, 23:39
nmohindru wrote:
iamcartic wrote:
Ans:5

8x6x4/3! = 32

Can you provide explanation please ... Thanks

I think i will also have a go at this.

Lets us first select 3 couples.. this can be done in C(4,3) ways = 4

Now out of these 3 couples chose one person each from each couple. This can be done in 2*2*2 ways = 8

Probability = 8*4 = 32
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Director
Joined: 27 May 2008
Posts: 543

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27 Jul 2008, 23:48
another method : calculate desired probablity and multiply it by total number of ways = 8C3 = 56

probablity of selecting first person = 8/8 (can be anyone)
probability of selecting secd person = 6/7 ( there are 7 left, but we can not take spouse of first person)
probability of selecting third person = 4/6 (there are 6 left but we can take only 4)

P = 8/8 * 6/7 * 4/6 = 4/7

fav ways = P * total ways = 4/7 * 56 = 32 Answer
Manager
Joined: 14 Jun 2008
Posts: 161

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28 Jul 2008, 00:47
Another view similiar to durgesh's

number of ways to choose first person : 8
number of ways to choose second person : 6 (cannot choose from the first couple)
number of ways to choose third person : 4 (cannot choose from the first and 2nd couple)

therefore, number of ways = 8 * 6 * 4

however, this is the number of permutations (as {couple 1 husband, couple 2 husband, couple 3 husband} same as {couple 2 husband, couple 1 husband, couple 3 husband} )

therefore, divide by 3!
therefore, number of ways = 8 * 6 * 4 / 3! = 32
Senior Manager
Joined: 19 Mar 2008
Posts: 351

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28 Jul 2008, 09:48
nmohindru wrote:
A committe of three people is to be chosen from four married couples . WHat is number of different committees that can be chosen if two people who are married to each other cannot both serve on committee

1. 16
2. 24
3. 26
4. 30
5. 32

Source -> GMAT prep

We have three positions to fill.
Choice of first position = 8 (let's call him Mr. Bush)
Choice of 2nd position = 6 (because we already chose Mr. Bush in first round and we cannot choose Mrs. Bush) , let call this 2nd committe member as Mr. Clinton.
Choice of 3rd position = 4 (because we Mr. Bush and Mr. Clinton are already in the committee, and we cannot choose Mrs Bush or Mrs. Clinton)

Combination is 8x6x4 = 192

what am I doing????
Manager
Joined: 15 Jul 2008
Posts: 206

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28 Jul 2008, 10:32
judokan wrote:
nmohindru wrote:
A committe of three people is to be chosen from four married couples . WHat is number of different committees that can be chosen if two people who are married to each other cannot both serve on committee

1. 16
2. 24
3. 26
4. 30
5. 32

Source -> GMAT prep

We have three positions to fill.
Choice of first position = 8 (let's call him Mr. Bush)
Choice of 2nd position = 6 (because we already chose Mr. Bush in first round and we cannot choose Mrs. Bush) , let call this 2nd committe member as Mr. Clinton.
Choice of 3rd position = 4 (because we Mr. Bush and Mr. Clinton are already in the committee, and we cannot choose Mrs Bush or Mrs. Clinton)

Combination is 8x6x4 = 192

what am I doing????

When u take 8x6x4, u r bringing arrangement (or order or position) into the expression. Since order of selection does not matter here, u need to divide 8x6x4 by the possible number of arrangements of 3 ppl. 3!=6. that will give u the true answer.
Senior Manager
Joined: 19 Mar 2008
Posts: 351

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28 Jul 2008, 10:39
bhushangiri wrote:
judokan wrote:
nmohindru wrote:
A committe of three people is to be chosen from four married couples . WHat is number of different committees that can be chosen if two people who are married to each other cannot both serve on committee

1. 16
2. 24
3. 26
4. 30
5. 32

Source -> GMAT prep

We have three positions to fill.
Choice of first position = 8 (let's call him Mr. Bush)
Choice of 2nd position = 6 (because we already chose Mr. Bush in first round and we cannot choose Mrs. Bush) , let call this 2nd committe member as Mr. Clinton.
Choice of 3rd position = 4 (because we Mr. Bush and Mr. Clinton are already in the committee, and we cannot choose Mrs Bush or Mrs. Clinton)

Combination is 8x6x4 = 192

what am I doing????

When u take 8x6x4, u r bringing arrangement (or order or position) into the expression. Since order of selection does not matter here, u need to divide 8x6x4 by the possible number of arrangements of 3 ppl. 3!=6. that will give u the true answer.

Thanks bhushangiri, but I think i only partly understand. Can you think of some examples?

Thanks again
Manager
Joined: 15 Jul 2008
Posts: 206

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28 Jul 2008, 10:59
1
KUDOS
Sure thing.. let us take a simple example..

Suppose u need to randomly select 2 out of 4 ppl.. A,B,C,D.

By your approach you would say, you have 4x3=12 ways.

But let us see...
BC, BD,
CD
are the only possible ways. ----- (1)

When u consider 4x3, you bring in the possibilities of
BA, CA, DA, CB, DB, and DC. When u have to simply select and not arrange, these are same as what is mentioned in (1).

a general point to remember.. when u are asked to "choose" or "select" then try and go with the combination approach i.e. using nCr formula. If you happen to use n(n-1)(n-2)... approach, then remember to divide by appropriate factorial term.

Last edited by bhushangiri on 28 Jul 2008, 11:34, edited 2 times in total.
Manager
Joined: 27 Mar 2008
Posts: 81

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28 Jul 2008, 11:26
GMBA85 wrote:
nmohindru wrote:
iamcartic wrote:
Ans:5

8x6x4/3! = 32

Can you provide explanation please ... Thanks

I think i will also have a go at this.

Lets us first select 3 couples.. this can be done in C(4,3) ways = 4

Now out of these 3 couples chose one person each from each couple. This can be done in 2*2*2 ways = 8

Probability = 8*4 = 32

I don't quite understand this part: Now out of these 3 couples chose one person each from each couple. This can be done in 2*2*2 ways = 8

If there are 3 couples and choose one person from each couple, wouldn't that be 6 people? What am I missing?
Senior Manager
Joined: 19 Mar 2008
Posts: 351

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28 Jul 2008, 22:24
bhushangiri wrote:
Sure thing.. let us take a simple example..

Suppose u need to randomly select 2 out of 4 ppl.. A,B,C,D.

By your approach you would say, you have 4x3=12 ways.

But let us see...
BC, BD,
CD
are the only possible ways. ----- (1)

When u consider 4x3, you bring in the possibilities of
BA, CA, DA, CB, DB, and DC. When u have to simply select and not arrange, these are same as what is mentioned in (1).

a general point to remember.. when u are asked to "choose" or "select" then try and go with the combination approach i.e. using nCr formula. If you happen to use n(n-1)(n-2)... approach, then remember to divide by appropriate factorial term.

Thanks, bhushangiri. +1 for you
Re: Couples   [#permalink] 28 Jul 2008, 22:24
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