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A committee of 3 people is to be chosen from four married

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Re: A committee of 3 people is to be chosen from four married  [#permalink]

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New post 12 Jan 2018, 14:52
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Hi All,

This question is fairly high-concept, but the math behind it isn't too complex. You have to be organized and you have to remember that we're forming GROUPS of people, so 'duplicate entries' are not allowed.

From the prompt, we have 4 married couples, but we have to form a group of 3 without putting a married couple in the group. Let's work through the logic in pieces...

The 1st person in the group can be ANY of the 8 people. Once we pick one of those 8, then we CANNOT pick the married partner to that person...

For the 2nd person in the group, we now have 6 people to choose from. Once we pick one of those 6, then we CANNOT pick the married partner to that person...

For the 3rd person in the group, we now have 4 people to choose from.

We now multiply those three numbers together: (8)(6)(4) = 192

We're NOT done though. Remember that we're forming GROUPS of people, and the 192 options we've figured out so far contain LOTS of duplicates... If you have 3 people: A, B and C, then that is just ONE group. However, in the above approach, you can end up with that group in 6 different ways:

ABC
ACB
BAC
BCA
CAB
CBA

Thus, every actual option has been counted 6 times, so we have to divide that total by 6...

192/6 = 32

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Re: A committee of 3 people is to be chosen from four married  [#permalink]

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New post 21 Feb 2018, 09:51
If there were no restrictions, number of ways in which 3 people can be chosen out of 8 people = C(8,3)

If we had a restriction that 2 of the 3 people necessarily belong to a couple, number of ways = C(4,1)*C(6,1)

So, number of ways in which the 3-people group will not have any couple = C(8,3) - C(4,1)*C(6,1) = 32
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Re: A committee of 3 people is to be chosen from four married  [#permalink]

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New post 02 Mar 2018, 22:05
So this is how i arrived at the answer. I need expert advice as to if my method is correct ??
Now there are 4 couples and so there are 8 members AB CD EF GH
So to form a team of 3 members without any condition there are 8c3 ways
So the answer is 56
Now from this if we deduct the number if ways in which the married couple are included we get the combination without the couple included.
So combination when married couple is included is there are 4 ways to select a married couple (Any 4 of the couples AB CD EF GH can be selected) and there is 6 ways to select the other remaining person (EX: If we select the couple AB then we can select CDEFG OR H ). therefore there are 6*4 ways = 24 ways
56-24=32 ways
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Re: A committee of 3 people is to be chosen from four married  [#permalink]

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New post 13 Mar 2018, 14:32
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LM wrote:
A committee of 3 people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A. 16
B. 24
C. 26
D. 30
E. 32


Take the task of selecting the 3 committee members and break it into stages.

Stage 1: Select the 3 couples from which we will select 1 spouse each.
There are 4 couples, and we must select 3 of them. Since the order in which we select the 3 couples does not matter, we can use COMBINATIONS
We can select 3 couples from 4 couples in 4C3 ways (4 ways)

Stage 2: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 3: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 4: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus create a 3-person committee) in (4)(2)(2)(2) ways (= 32 ways)

Answer = E

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Re: A committee of 3 people is to be chosen from four married  [#permalink]

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New post 13 Mar 2018, 17:07
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Four couples: AB CD EF GH
3 positions to fill

Total to be chosen is 8C3 = 56

If the first two people on the committee are A&B, then the 3rd can be any of the 6 remaining couples. If C&D are picked then the remaining people to choose from are 6. This continues for E&F and G&H.

6+6+6+6 = 24 unwanted events.

56-24 = 32

Answer E
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Re: A committee of 3 people is to be chosen from four married  [#permalink]

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New post 15 Apr 2018, 13:30
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Concept: Total selection without the constraint = Total ways of selection - Total ways of selection with constraint.

1) We can choose 3 people out of 8 (4 couples, so total=4*2=8) in \(8C3\) ways.
2) We can choose 1 couple out of 4 in 4C1. Note, a couple is 2 people so two places out of 3 from the committee is taken.
3) We can choose 1 person out of 6 [8- couple(2)=8-2=6] in \(6C1\).

Using concept, Total ways of making the committee without couple/constraint = \(8C3\)-(\(4C1\)+\(6C1\))=56-24=32 (E).
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Re: A committee of 3 people is to be chosen from four married  [#permalink]

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New post 12 May 2018, 17:09
You are trying to form a committee of six people at work to run a group project. If you have fifteen coworkers to choose from, how many different committees can you form?
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Re: A committee of 3 people is to be chosen from four married  [#permalink]

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New post 13 May 2018, 00:47
marycosme2005 wrote:
You are trying to form a committee of six people at work to run a group project. If you have fifteen coworkers to choose from, how many different committees can you form?


Choosing 6 out of 15: \(15C6 = \frac{15!}{6!(15-6)!}=\frac{15!}{6!9!}=5,005\).

21. Combinatorics/Counting Methods



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Re: A committee of 3 people is to be chosen from four married  [#permalink]

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New post 14 May 2018, 12:38
Total couples: 4
Available positions: 3

Select 3 couples : 4C3 = 4

Now three couples and we have 3 positions and one from each of the couple and can take the position.

Hence total arrangements: 6*4*2 = 48

Since we are just selecting and not arranging, we 'unarrange' : 48/3! = 8

Total possible selections: 8*4 = 32
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A committee of 3 people is to be chosen from four married  [#permalink]

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New post 28 May 2018, 13:35
Bunuel wrote:
A committee of 3 people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?
A. 16
B. 24
C. 26
D. 30
E. 32

One of the approaches:

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) out of 4 to send only one "representatives" to the committee: 4C3=4.

But each of these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.


Total # of ways: 4C3*2^3=32.

Answer: E.



Hey Bunuel

is my apprach correct

total # of ways 8C3 = 56

there are 4 couples / so i will take one couple (two persons) as one unit - choosing the 1 couple from 4 --> 4C1 = 4

now i am left with 3 couples which means 6 people are left so i need to choose one more person from 6 ---> 6C1 = 6

total number of possible groups in which two pearsons married to eaach other + someone else 6C1*4C1 ---> 6*4 = 24

56-24 = 32

Answer E: 32

pushpitkc hi there :) please advise if my approach is correct ? :-) many thanks :)
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Re: A committee of 3 people is to be chosen from four married  [#permalink]

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New post 29 May 2018, 01:52
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dave13 wrote:
Bunuel wrote:
A committee of 3 people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?
A. 16
B. 24
C. 26
D. 30
E. 32

One of the approaches:

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) out of 4 to send only one "representatives" to the committee: 4C3=4.

But each of these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.


Total # of ways: 4C3*2^3=32.

Answer: E.



Hey Bunuel

is my apprach correct

total # of ways 8C3 = 56

there are 4 couples / so i will take one couple (two persons) as one unit - choosing the 1 couple from 4 --> 4C1 = 4

now i am left with 3 couples which means 6 people are left so i need to choose one more person from 6 ---> 6C1 = 6

total number of possible groups in which two pearsons married to eaach other + someone else 6C1*4C1 ---> 6*4 = 24

56-24 = 32

Answer E: 32

pushpitkc hi there :) please advise if my approach is correct ? :-) many thanks :)


Hi dave13 - This method is absolutely correct.
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Re: A committee of 3 people is to be chosen from four married &nbs [#permalink] 29 May 2018, 01:52

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