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

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

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21 Jul 2013, 19:38
AbeinOhio wrote:
I'm pretty lost on this one..

Here was the logic that i thought would work but obviously I am missing something:

5 couples = 10 people

So first position there are 10 options

Second position, the first person's mate is excluded now so instead of having 9 options you have 8 options

3rd position the first and 2nd person's mates are both excluded now so instead of having 7 options you have 6 options

so I got 10 * 9 * 8 = 480

What am I missing?

What if you started out by choosing the first person in 1 one way instead of 10.

So you pick a person of the 10, doesn't matter who. For the next slot you have 8 choices. and the next 6. so (1)(8)(6). Why isn't this approach valid?
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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21 Jul 2013, 22:05
alphabeta1234 wrote:
AbeinOhio wrote:
I'm pretty lost on this one..

Here was the logic that i thought would work but obviously I am missing something:

5 couples = 10 people

So first position there are 10 options

Second position, the first person's mate is excluded now so instead of having 9 options you have 8 options

3rd position the first and 2nd person's mates are both excluded now so instead of having 7 options you have 6 options

so I got 10 * 9 * 8 = 480

What am I missing?

What if you started out by choosing the first person in 1 one way instead of 10.

So you pick a person of the 10, doesn't matter who. For the next slot you have 8 choices. and the next 6. so (1)(8)(6). Why isn't this approach valid?

Why do you have 1 choice and not 10?
Why do you have 8 choices for the second pick and not 9?
Why do you have 6 choices for the third pick and not 8?

The reason why AbeinOhio's solution is not correct is explained here: if-a-committee-of-3-people-is-to-be-selected-from-among-98533.html#p1051830

Hope it helps.
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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21 Aug 2013, 17:40

5C1*2C1* 4C1*2C1* 8C1=640, being:

5C1*2C1 - ways of choosing one couple out of 5, multiplied by 2C1 ways of choosing one person of a couple
4C1*2C1 - ways of choosing another couple out of 4, multiplied by 2C1 ways of choosing one person of a couple
8C1 - people that can occupy the third spot
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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21 Aug 2013, 22:58
brunawang wrote:

5C1*2C1* 4C1*2C1* 8C1=640, being:

5C1*2C1 - ways of choosing one couple out of 5, multiplied by 2C1 ways of choosing one person of a couple
4C1*2C1 - ways of choosing another couple out of 4, multiplied by 2C1 ways of choosing one person of a couple
8C1 - people that can occupy the third spot

Ok, here is what is wrong with your solution.
Say, the couples are (A1, A2), (B1, B2), (C1, C2), (D1, D2) and (E1, E2)

Now you cannot have 2 people from the same couple.

Two different scenarios in your solution:

You select one couple in 5 ways. Say you selected (C1, C2). In two ways you selected one of them. You got C2.
You select one couple in 4 ways now. Say you selected ((E1, E2). In two ways you selected one of them. You got E2.
You selected one person out of 8 in 8 ways, You got A2.

You select one couple in 5 ways. Say you selected (E1, E2). In two ways you selected one of them. You got E2.
You select one couple in 4 ways now. Say you selected ((C1, C2). In two ways you selected one of them. You got C2.
You selected one person out of 8 in 8 ways, You got A2.

Notice that they give you the same team but you have counted these two as different selections. Hence your answer is incorrect.
When making a selection, try to use 5C3 method. It helps you think clearly. You select 3 couples out of the 5. Now from each couple you select one person out of the two. So you get 5C3*2*2*2 - there is no double counting here.
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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25 Oct 2013, 19:01
8
Using slot method:
First person can be chosen -> 10 ways,
2nd person can be chosen -> 8 ways (1st person and his wife are not candidates) and
3rd person can be chosen -> 6 ways (1st person/2nd person and their wives are out)
Answer -> 10*8*6/6 = 80 (divide by 6 because the 3 people can be chosen in any order (i.e. 3! = 3*2*1 ways))
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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10 May 2014, 11:14
1
VeritasPrepKarishma wrote:
brunawang wrote:

5C1*2C1* 4C1*2C1* 8C1=640, being:

5C1*2C1 - ways of choosing one couple out of 5, multiplied by 2C1 ways of choosing one person of a couple
4C1*2C1 - ways of choosing another couple out of 4, multiplied by 2C1 ways of choosing one person of a couple
8C1 - people that can occupy the third spot

Ok, here is what is wrong with your solution.
Say, the couples are (A1, A2), (B1, B2), (C1, C2), (D1, D2) and (E1, E2)

Now you cannot have 2 people from the same couple.

Two different scenarios in your solution:

You select one couple in 5 ways. Say you selected (C1, C2). In two ways you selected one of them. You got C2.
You select one couple in 4 ways now. Say you selected ((E1, E2). In two ways you selected one of them. You got E2.
You selected one person out of 8 in 8 ways, You got A2.

You select one couple in 5 ways. Say you selected (E1, E2). In two ways you selected one of them. You got E2.
You select one couple in 4 ways now. Say you selected ((C1, C2). In two ways you selected one of them. You got C2.
You selected one person out of 8 in 8 ways, You got A2.

Notice that they give you the same team but you have counted these two as different selections. Hence your answer is incorrect.
When making a selection, try to use 5C3 method. It helps you think clearly. You select 3 couples out of the 5. Now from each couple you select one person out of the two. So you get 5C3*2*2*2 - there is no double counting here.

Dear Karishma

So in selection of pair qs; we can select using 10*8*6 approach but remember to divide it by 3! ways?

same could be applied to any selection and couple qs???
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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12 May 2014, 20:19
nandinigaur wrote:
Dear Karishma

So in selection of pair qs; we can select using 10*8*6 approach but remember to divide it by 3! ways?

same could be applied to any selection and couple qs???

Yes, when you need to use combination (i.e. selection only), you can use the basic counting principle (i.e. the 10*8*6 approach), but you must un-arrange at the end by dividing by n!.

Check out this post: http://www.veritasprep.com/blog/2011/11 ... binations/

In this, I have discussed how to use the basic counting principle for combination with examples. It will help you out.
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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22 May 2014, 22:26
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1
Hi achakrav2694,

Ordering is not Required in Selection.

We can have just 4 Cases:

1) Selecting All 3 husbands: This would be$$5C3 = 10$$
2) Selecting 2 husbands and 1 Wife: $$5C2 * 3$$ (As Wife cannot be for the 2 husbands selected) $$= 30$$
3) Selecting All 3Wives: This would be $$5C3 = 10$$
4) Selecting 2 Wives and 1 Husband: $$5C2 * 3$$(As Husband cannot be for the 2 Wives selected) $$= 30$$

Total Commitees $$= 10 + 30 + 10 + 30 = 80$$

Rgds,
Rajat
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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15 Jul 2014, 04:06
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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.

wow superb approach I would give you 100 Kudos. THANKS Bunuel.
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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15 Sep 2014, 18:02
2
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.
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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26 Aug 2016, 00:47
1
3
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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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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Updated on: 27 Aug 2016, 06:02
4
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 married coup  [#permalink]

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30 Aug 2016, 07:36
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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 married coup  [#permalink]

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19 Jan 2017, 18:24
Maxirosario2012 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?

(1) Combinatorial approach:

$$C^5_3*(C^2_1)^3 =\frac{5*4*3}{3*2}* 2^3 = 10*8 = 80$$

(2) Reversal combinatorial approach:

Total number of groups: $$C^10_3 = 120$$
Total number of groups with married people: $$C^5_1 * C^4_1 * C^2_1 = 5*4*2 = 40$$

120 - 40 = 80

Hello, I have a question when you calculate the No of groups of married people you use C^2_1 but i saw a similar problem with that approach which uses C2^2
this one:If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it? when we exclude the committees with married ppl
can you explain me the difference? thank you in advance!
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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23 Jan 2017, 01:14
1
Alexangeo wrote:
Maxirosario2012 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?

(1) Combinatorial approach:

$$C^5_3*(C^2_1)^3 =\frac{5*4*3}{3*2}* 2^3 = 10*8 = 80$$

(2) Reversal combinatorial approach:

Total number of groups: $$C^10_3 = 120$$
Total number of groups with married people: $$C^5_1 * C^4_1 * C^2_1 = 5*4*2 = 40$$

120 - 40 = 80

Hello, I have a question when you calculate the No of groups of married people you use C^2_1 but i saw a similar problem with that approach which uses C2^2
this one:If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it? when we exclude the committees with married ppl
can you explain me the difference? thank you in advance!

You have 5 couples. First you select 3 couples. This is done in 5C3 ways (select 3 out of 5)
Next out of these 3 selected couples, from each couple, select 1 of the 2 people in 2C1 ways. Since we do it for each couple, we get 2C1 * 2C1 * 2C1

Total ways = 5C3 * 2C1 * 2C1 * 2C1 ways

If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

The same method will be used for this question as well. If there is something else done here in the solution, please send me the link of this question.
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Re: If a committee of 3 people is to be selected from among 5 married coup  [#permalink]

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25 Jan 2017, 10:18
kwhitejr 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.

If there are 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 be selected for the committee. This reduces to 8 the number of remaining people from which to select the second person (one person has already been selected, and that person’s spouse 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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Re: If a committee of 3 people is to be selected from among 5 married coup  [#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 married coup  [#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 married coup  [#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 married coup  [#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 1374 times ]

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Re: If a committee of 3 people is to be selected from among 5 married coup   [#permalink] 29 Apr 2018, 16:03

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