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If a committee of 4 people is to be selected from among 6 ma

by gmatter2012 » Wed Sep 12, 2012 4:14 am
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)

P.S Good question to clear certain concepts.
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by Brent@GMATPrepNow » Wed Sep 12, 2012 6:33 am
Can you please limit your posts to one question per thread? Otherwise, the conversations can get pretty confusing.
Also, can you provide the answer choices?
In some cases, we can use the answer choices to direct our approach (or our guesses)

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by coolhabhi » Wed Sep 12, 2012 6:36 am
gmatter2012 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)

Answer : [spoiler]12C1 * 10C1 * 8C1 * 6C1 = 12 * 10 * 8 * 6[/spoiler]
Suppose the couples are (a,b), (c,d), (e,f), (g,h), (i,j), (k,l)
Ways to choose "None of them should be married to each other" = [spoiler]12 * 10 * 8 * 6[/spoiler]
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by coolhabhi » Wed Sep 12, 2012 6:39 am
gmatter2012 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)
Answer : [spoiler]12C1 * 1 * 10C1 * 8C1 = 12 * 1 * 10 * 8[/spoiler]
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by coolhabhi » Wed Sep 12, 2012 6:40 am
gmatter2012 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)
Answer : [spoiler]12C1 * 1 * 10C1 * 1 = 12 * 1 * 10 * 1[/spoiler]
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by Brent@GMATPrepNow » Wed Sep 12, 2012 6:43 am
gmatter2012 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)
Here's one approach.


Aside: Notice that we must select 1 spouse from 4 of the 6 couples.

Take the task of selecting the 4 members and break it into stages.

Stage 1: Select the 4 couples from which we will select 1 spouse each.
There are 6 couples, and we must select 4. Since the order in which we select the 4 couples does not matter, this stage can be accomplished in 6C4 ways (15 ways)

Stage 2: Take one of the 4 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 4 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 4 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 5: Take one of the 4 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 5 stages (and thus create a 4-person committee) in (15)(2)(2)(2)(2) ways (= 240 ways)

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775
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by gmatter2012 » Thu Sep 13, 2012 8:44 pm
Brent@GMATPrepNow wrote:
gmatter2012 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)
Here's one approach.


Aside: Notice that we must select 1 spouse from 4 of the 6 couples.

Take the task of selecting the 4 members and break it into stages.

Stage 1: Select the 4 couples from which we will select 1 spouse each.
There are 6 couples, and we must select 4. Since the order in which we select the 4 couples does not matter, this stage can be accomplished in 6C4 ways (15 ways)

Stage 2: Take one of the 4 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 4 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 4 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 5: Take one of the 4 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 5 stages (and thus create a 4-person committee) in (15)(2)(2)(2)(2) ways (= 240 ways)

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775


Brent this is fine
However I was used to selecting the total number of ways 4 people can be selected and then subtracting the ways
one or two couple can selected

10C4 number of selecting 4 from 10 = 495

now ways to select exactly one couple ??

ways to select exactly two couples ??

495- ( ways to select one couple + ways to select two couples)= ways to select no couple

if you could just show , the logic behind selecting exactly one couple and exactly two couple that would really help.

Thank you.
Last edited by gmatter2012 on Sun Aug 11, 2013 10:15 pm, edited 1 time in total.
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by Anurag@Gurome » Thu Sep 13, 2012 9:21 pm
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)
This means 1 couple and 2 non-couple people.
for selecting 1 couple we have 6C1 options.
as we have selected 1 couple already, there are 5 couples left.
in these 5 couples we need to have 2 couples to select and 1 person from each couple.
so it will be 5C2 * 2 * 2.
ans = 6c1 * 5c2 * 2 * 2 = 240

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)
It is same as selecting 2 couples from 6 couples = 6C2 = 15
However I was used to select the total number of ways 4 people can be selected and then subtracting the ways
one or two couple can selected
Number of ways of selecting 4 people from 6 couples (12 people) = 12C4 = 495
the number of ways to select 4 people who are non-couple = total number of ways - number of ways 1 couple and 2 non-couple - number of ways 2 couples
= 495 - 240 -15 = 240
But the solution Brent gave is the simpler one.
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by gmatter2012 » Thu Sep 13, 2012 11:59 pm
Anurag@Gurome wrote: Number of ways of selecting 4 people from 6 couples (12 people) = 12C4 = 495
the number of ways to select 4 people who are non-couple = total number of ways - number of ways 1 couple and 2 non-couple - number of ways 2 couples
= 495 - 240 -15 = 240
But the solution Brent gave is the simpler one.
agreed that Brents solution is a simpler one , but that's ok when we have to select members such that all are singles ( no couples )

here we use 6c4*(2)^4 = 240 ways , where all are singles

now what I was looking for is this

exactly one couple = 12*1*10*8/2!2!= 240

or

6C1 * (10 *8/2!) = 240


for two couples

6 *5/2! = 15 ( first we have 6 couples , then we have 5 couples to choose from dividing by 2! to show that order is not important )

however I am having difficulty using single people rather than couple

12*1*10*1= this goes in the numerator for no. of ways to select 2 couples , right? what is the logic for the denominator .

Seeing one couple case I would put 2!2! in the denominator , but that doesn't work?
Last edited by gmatter2012 on Sun Aug 11, 2013 10:34 pm, edited 1 time in total.
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by Anurag@Gurome » Fri Sep 14, 2012 2:56 am
for two couples

6 *5/2! = 15 ( first we have 6 couples , then we have 5 couples to choose from dividing by 2! to show that order is not important )

however I am having difficulty using using single people rather than couple

12*1*10*1= this goes in the numerator for no. of ways to select 2 couples , right? what is the logic for the denominator .

Seeing one couple case I would put 2!2! in the denominator , but that doesn't work?
lets take the couples be (x1,y1)....(x6,y6)
In the numerator 12*1*10*1
lets take the 1st time you have chosen x1. so y1 will be chosen. the next time you have chosen x2. so y2 will be chosen.
what have you chosen is x1,y1,x2,y2

now you have chosen 1st time y1. so you are choosing x1. the next time you have chosen x2. so you are choosing y2.
you have chosen y1,x1,x2,y2
we are dividing by 2 because we are counting twice the selection.

now you have chosen 1st time x1. so you are choosing y1. the next time you have chosen y2. so you are choosing x2.
you have chosen x1,y1,y2,x2
we are dividing again by 2 because we are counting twice the selection.

now you have chosen 1st time x1. so you are choosing y2. the next time you have chosen x1. so you are choosing y1.
you have chosen x2,y2,x1,y1
are you not counting the same selection again? You have forgot to consider choosing one couple in the 1st time or 2nd time results in same selection. So it should be divided by 2.
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by gmatter2012 » Fri Sep 14, 2012 4:17 am
Anurag@Gurome wrote:
lets take the couples be (x1,y1)....(x6,y6)
In the numerator 12*1*10*1
lets take the 1st time you have chosen x1. so y1 will be chosen. the next time you have chosen x2. so y2 will be chosen.
what have you chosen is x1,y1,x2,y2 .........
so basically you are saying that 12*1*10*1 needs to be divided by an extra 2!
so what we have is this 12*1*10*1/2!2!2! = 15 am I correct ?

Clearly this method is not ideal
after careful analysis I find the best way to solve such cases is as you have shown earlier

Thank you .. 6C1*5C2 *2 *2 = 240 This is an awesome way .

Just to test my understanding below is a case which I have developed , this covers all cases and hopefully will help others too.

Suppose there are six couples , and six members need to be chosen ,Find ?

1)No. of ways exactly one couple can be selected
2) No. of ways exactly two couples can be selected
3)ways exactly 3 couples can be selected
4) ways such all are singles , i.e no couple is chosen?


one couple
6C1 *5C4 *(2)^4= 480

two couple
6C2*4C2*(2)^2= 360

Three couple
6C3= 20

No couple
6C6 * (2)^6= 64

** ** ** *** **


Suppose there are Five couples and we need to select 4, how to find?

1)No. of ways exactly one couple can be selected
2) No. of ways exactly two couples can be selected
3) ways such all are singles , i.e no couple is chosen?

One couple only

5C1 *4C2*(2)^2= 120

two couple only

5C2= 10


All singles

5C4 *(2)4 = 5 *16= 80

Hope these are correct?
Last edited by gmatter2012 on Sun Aug 11, 2013 11:28 pm, edited 1 time in total.
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by Anurag@Gurome » Fri Sep 14, 2012 9:31 am
so basically you are saying that 12*1*10*1 needs to be divided by an extra 2!
so what we have is this 12*1*10*1/2!2!2! = 15 am I correct ?
yes, you are right.

Suppose there are six couples , and six members need to be chosen ,Find ?

1)No. of ways exactly one couple can be selected
2) No. of ways exactly two couples can be selected
3)ways exactly 3 couples can be selected
4) ways such all are singles , i.e no couple is chosen?


one couple
6C1 *5C4 *(2)^4= 480

two couple
6C2*4C2*(2)^2= 360

Three couple
6C3= 20

No couple
6C6 * (2)^6= 64

** ** ** *** **


Suppose there are Five couples and we need to select 4, how to find?

1)No. of ways exactly one couple can be selected
2) No. of ways exactly two couples can be selected
3) ways such all are singles , i.e no couple is chosen?

One couple only

5C1 *4C2*(2)^2= 120

two couple only

5C2= 10


All singles

5C4 *(2)4 = 5 *16= 80

Hope these are correct?
The answers look good.
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by gmatter2012 » Fri Sep 14, 2012 3:30 pm
Anurag@Gurome wrote:
so basically you are saying that 12*1*10*1 needs to be divided by an extra 2!
so what we have is this 12*1*10*1/2!2!2! = 15 am I correct ?
yes, you are right.

Suppose there are six couples , and six members need to be chosen ,Find ?

1)No. of ways exactly one couple can be selected
2) No. of ways exactly two couples can be selected
3)ways exactly 3 couples can be selected
4) ways such all are singles , i.e no couple is chosen?


one couple
6C1 *5C4 *(2)^4= 480

two couple
6C2*4C2*(2)^2= 360

Three couple
6C3= 20

No couple
6C6 * (2)^6= 64

** ** ** *** **


Suppose there are Five couples and we need to select 4, how to find?

1)No. of ways exactly one couple can be selected
2) No. of ways exactly two couples can be selected
3) ways such all are singles , i.e no couple is chosen?

One couple only

5C1 *4C2*(2)^2= 120

two couple only

5C2= 10


All singles

5C4 *(2)4 = 5 *16= 80

Hope these are correct?
The answers look good.
Full Credit goes to you.That was an Awesome method taught by you.Means a lot to me.

(Thank you)^100
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