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# If a committee of 3 people is to be selected from among 5

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If a committee of 3 people is to be selected from among 5  [#permalink]

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26 Aug 2016, 00:47
2
I will tell u easy way of doing --always keep it simple rather than using terminologies

total 5 men and 5 women are there.

committee of 3 people is to be selected....so it can be in following ways..

1.MMM -- all 3 men in commitee , so 5C3 = 10 ways
2.FFF ---all 3 women in commitee s0 5c3 = 10ways
3.MMF-- two men in committee 5C2 ways and 1female ( be careful here(only Trap)1 women is to be selected from 3 women not 5 women)as
2 women will be wife of 2 men already selected and both husband wife cannot be in committee. Note order of selection doesn't matter here.
5C2 * 3C1 = 30 ways

4.FFM ---two women in committee 5C2 ways and men( be careful here) 1 men is to be selected from 3 men not 5 men,as
2 men are husbands of 2 women already selected and both husband wife cannot be in committee
5C2 * 3C1 = 30 ways

total ways =10+10+30 +30
= 80
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If a committee of 3 people is to be selected from among 5  [#permalink]

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Updated on: 27 Aug 2016, 06:02
3
kirankp wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

We are given that there are five married couples (or 10 people) and we need to determine the number of ways of choosing 3 people in which no two people are a married couple. So this is a special combination problem. Before we tackle this problem, let’s review a combination problem with no restrictions.

With no restrictions, the number of ways of choosing 3 people from 10 people is 10C3, which is calculated as follows:

(10 x 9 x 8)/3! = 120

10, 9 and 8, in the numerator, represent the number of ways the first, second and third person can be chosen respectively. We divide the numerator by 3! because in a combination problem we do not care about the order in which the 3 people are chosen.

However, in this (special combination) problem, 3 people are chosen in which no married couple can serve together on the committee. The first person could be any one of the 10 people. However, once a person is selected, that person’s spouse cannot also be selected for the committee. This reduces the second person to 8 possible people (one person has already been selected, and that person’s spouse now cannot be selected). Once the second person is chosen for the committee, that person’s spouse cannot be chosen. This reduces the number of people who could be chosen as the third person to 6. Therefore, the number of ways of choosing these 3 people is:

(10 x 8 x 6)/3! = 80

Thus, there are 80 ways to choose such a committee.

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Originally posted by ScottTargetTestPrep on 26 Aug 2016, 09:53.
Last edited by ScottTargetTestPrep on 27 Aug 2016, 06:02, edited 2 times in total.
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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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27 Aug 2016, 05:19
1
ScottTargetTestPrep wrote:
kirankp wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

We are given that there are four married couples (or 8 people) and we need to determine the number of ways of choosing 3 people in which no two people are a married couple. So this is a special combination problem. Before we tackle this problem, let’s review a combination problem with no restrictions.

With no restrictions, the number of ways of choosing 3 people from 8 people is 8C3, which is calculated as follows:

(8 x 7 x 6)/3! = 56

8, 7 and 6, in the numerator, represent the number of ways the first, second and third person can be chosen respectively. We divide the numerator by 3! because in a combination problem we do not care about the order in which the 3 people are chosen.

However, in this (special combination) problem, 3 people are chosen in which no married couple can serve together on the committee. The first person could be any one of the 8 people. However, once a person is selected, that person’s spouse cannot also be selected for the committee. This reduces the second person to 6 possible people (one person has already been selected, and that person’s spouse now cannot be selected). Once the second person is chosen for the committee, that person’s spouse cannot be chosen. This reduces the number of people who could be chosen as the third person to 4. Therefore, the number of ways of choosing these 3 people is:

(8 x 6 x 4)/3! = 32

Thus, there are 32 ways to choose such a committee.

Bro, we are given 5 married couple not 4. Please check your answer.
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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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30 Aug 2016, 07:36
Top Contributor
1
kirankp wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

Take the task of creating a committee and break it into stages.

Stage 1: Select 3 COUPLES
Since the order in which we select the couples does not matter, we can use COMBINATIONS
We can select 3 couples from 5 couples in 5C3 ways ( = 10 ways)

ASIDE: If anyone is interested, we have a video on calculating combinations (like 5C3) in your head (see bottom of post)

At this point, we have selected 3 COUPLES, which we'll call A, B and C. We're now going to select ONE person from each couple to be on the committee.

Stage 2: Select 1 person from couple A
There are 2 people in this couple, so we can complete this stage in 2 ways.

Stage 3: Select 1 person from couple B
There are 2 people in this couple, so we can complete this stage in 2 ways.

Stage 4: Select 1 person from couple C
There are 2 people in this couple, so we can complete this stage in 2 ways.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create a 3-person committee) in (10)(2)(2)(2) ways (= 80ways)

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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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20 Mar 2017, 10:20
xcusemeplz2009 wrote:
kirankp wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
A. 20
B. 40
C. 50
D. 80
E. 120

total no for selecting 3 out of 10=10c3=120

no. of ways in which no two married people included= tot- 2 married couple included
2 married couple can be included in 5c1( no. of ways selecting a couple) * 8c1( no. of ways selecting the third person)=5 * 8=40

reqd comb=120-40=80
hence D

I am very weak in Probability and P&C so sorry I couldn't understand the 8c1 part:c . Would Love your help on it
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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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20 Mar 2017, 10:22
Bunuel wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

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) to send only one "representatives" to the committee: 5C3=10.

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

Total # of ways: 5C3*2^3=80.

Bunuel Can you please explain the 2^3 part, I am very weak in P&C.
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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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20 Mar 2017, 10:40
jamescath wrote:
Bunuel wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

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) to send only one "representatives" to the committee: 5C3=10.

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

Total # of ways: 5C3*2^3=80.

Bunuel Can you please explain the 2^3 part, I am very weak in P&C.

With this approach we are choosing couples who then will delegate on person per couple to the committee. Now, there will be 3 couples selected. Each can delegate either husband or wife, thus 2 choices per couple. For 3 couples 2*2*2 = 8 choices.

Hope it's clear.
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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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20 Mar 2017, 11:47
Bunuel wrote:
jamescath wrote:
Bunuel wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

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) to send only one "representatives" to the committee: 5C3=10.

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

Total # of ways: 5C3*2^3=80.

Bunuel Can you please explain the 2^3 part, I am very weak in P&C.

With this approach we are choosing couples who then will delegate on person per couple to the committee. Now, there will be 3 couples selected. Each can delegate either husband or wife, thus 2 choices per couple. For 3 couples 2*2*2 = 8 choices.

Hope it's clear.

Yes it is:) Thanks once again
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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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21 Mar 2017, 03:39
kirankp wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

selection for 10 people = 10C3
slection with one couple = 5C1. 8C1

therefore total ways = 10C3 - 5C1.8C1 = 120 - 40 = 80
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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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05 Apr 2017, 01:13
We can only choose either the husband or the wife from each couple, so this becomes 2C1. Like this we have to chose 3 people, so 2c1 x 2c1 x 2c1. And since we have a total of 5 couples we need to choose 3 out of 5, which is 5c3.

(2C1x2C1x2C1) * (5C3) = 8 x 2 = 80 ways
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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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22 Apr 2017, 18:36
Bunuel wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

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) to send only one "representatives" to the committee: 5C3=10.

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

Total # of ways: 5C3*2^3=80.

Bunuel One thing that's kind of confusing me is restrictions - so generally 10c3 is the total number of arrangements do we basically subtract the total number of arrangments (arrangements that include arrangements that violate the restriction) by the number of arrangements that violate the restriction in order to calculate the number of arrangments that conform to the restriction?
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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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25 Jul 2017, 07:21
Explanation:

To select 3 people from 5 married couples so that no couple is selected, we need to select these 3 people from 3 different couples selecting one from each.

Number of ways to select 3 couples from 5 couples = 5C3
Number of ways to select 1 person from a couple = 2C1
Now, we have to select 3 couples and 1 person from each couple, i.e. 5C3 × (2C1 × 2C1 × 2C1) = 80.

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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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24 Oct 2017, 20:06
1
cabelk wrote:
"If a committee of 3 people is to be selected"
Combo box arrangement
(_)(_)(_)/3!

"from among 5 married couples"
Bag of 10 choices: A,B,C,D,E,F,G,H,I,J

"so that the committee does not include two people who are married to each other"
First slot has 10 choices
(10)(_)(_)/3!

but the choice eliminates the spouse. The second slot has 8 choices
(10)(8)(_)/3!

but the choice eliminates another spouse. The third slot has 6 choices
(10)(8)(6)/3!

"how many such committees are possible?"
(10)(8)(6)/(3*2) = 80

I am not able to understand this conceptually but I think I can memorize it
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Re: If a committee of 3 people is to be selected from among 5  [#permalink]

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29 Oct 2017, 10:13
JackSparr0w wrote:
For those as challenged with math as myself, try this (apologies if this has already been done):

Consider the "slot method"

We need to select three people out a total of ten people (5 couples=10ppl), so create three "slots"

_ _ _
1 2 3

The only restrictions we have are that we cant pick an individual and that individual's spouse, nor can we select any individual twice. Knowing this, we can put any of our 10 people in the first slot. The second spot, however, will be limited to 8ppl (10 total less the person we placed in spot one, leaves 9, less that person's spouse, leaves 8). The third slot will only have 6 possibilities (10 total less the two ppl already placed and each of their spouses). Fill in the slots:

10 8 6
1 2 3

We would then multiply across to get 10*8*6=480

When using the Slot method, if the order of the selections does not matter (as is the case here) we must divide the the product of the slots by n!, where n is just the number of slots.

480/3! =480/6 = 80

Generally, you'd want to simplify before dividing, so the 2*3 in the 3! term would cancel the 6 in the numerator, and you'd just be left with (10*8)/1 or 80.

Thanks for the explanation, I think I am starting to get it. Could you please try to explain the passage in bold? I understand that there is the need to divide by something to get to the actual result, but how do you come up with 6?

Thanks a lot!
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If a committee of 3 people is to be selected from among 5  [#permalink]

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29 Apr 2018, 16:03
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Attached is a visual that should help.

-Brian
Attachments

Screen Shot 2018-04-29 at 4.03.28 PM.png [ 790.77 KiB | Viewed 514 times ]

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