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555-605 (Medium)|   Combinations|            
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KarishmaB
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If a committee of 3 people is to be selected from among 5 married coup 21 Aug 2013, 22:58
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brunawang
Can someone please help me? I don't know what I am doing wrong.

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
Show more

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.
Your team (A2, C2, E2)

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.
Your team (A2, C2, E2)

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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If a committee of 3 people is to be selected from among 5 married coup 25 Oct 2013, 19:01
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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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If a committee of 3 people is to be selected from among 5 married coup 10 May 2014, 11:14
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brunawang
Can someone please help me? I don't know what I am doing wrong.

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.
Your team (A2, C2, E2)

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.
Your team (A2, C2, E2)

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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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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If a committee of 3 people is to be selected from among 5 married coup Updated on: 17 Oct 2022, 00:43
Originally posted by KarishmaB on 12 May 2014, 20:19.
Last edited by KarishmaB on 17 Oct 2022, 00:43, edited 1 time in total.
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nandinigaur

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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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!.
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If a committee of 3 people is to be selected from among 5 married coup 22 May 2014, 22:26
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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 \(= 10 + 30 + 10 + 30 = 80\)

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If a committee of 3 people is to be selected from among 5 married coup 15 Jul 2014, 04:06
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Bunuel
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

Another way to think about this problem:

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.

Answer: D.
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wow superb approach I would give you 100 Kudos. THANKS Bunuel.
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If a committee of 3 people is to be selected from among 5 married coup 15 Sep 2014, 18:02
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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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If a committee of 3 people is to be selected from among 5 married coup 26 Aug 2016, 00:47
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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 ( :twisted: 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 :lol: :P
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If a committee of 3 people is to be selected from among 5 married coup Updated on: 27 Aug 2016, 06:02
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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kirankp
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
Show more


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.

Answer: D
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If a committee of 3 people is to be selected from among 5 married coup 30 Aug 2016, 07:36
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kirankp
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
Show more

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)

Answer:
Show Spoiler
D

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If a committee of 3 people is to be selected from among 5 married coup 19 Jan 2017, 18:24
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Maxirosario2012
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
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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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If a committee of 3 people is to be selected from among 5 married coup 23 Jan 2017, 01:14
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Alexangeo
Maxirosario2012
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!
Show more

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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If a committee of 3 people is to be selected from among 5 married coup 20 Mar 2017, 10:22
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Bunuel
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

Another way to think about this problem:

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.

Answer: D.
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Bunuel Can you please explain the 2^3 part, I am very weak in P&C.
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If a committee of 3 people is to be selected from among 5 married coup 20 Mar 2017, 10:40
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Bunuel
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

Another way to think about this problem:

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.

Answer: D.

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

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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If a committee of 3 people is to be selected from among 5 married coup 18 Jul 2021, 03:46
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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?

You have to select a committee of 3 people. It means A, B, C form the same committee as B, C, A i.e. you don't have to arrange people. But when you say 'first person' can be selected in 10 ways, you are arranging the 3 people in first, second and third spots. So according to you, A, B, C and B, C, A are different committees.
So all you need to do in un-arrange. To arrange 3 people, you multiply by 3!
To un-arrange, you will divide by 3!
480/3! = 80

There's your answer.

Check out this post for a detailed explanation:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/11 ... nstraints/
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VeritasKarishma any idea why my approach didnt work


total number of ways: 5C3 = 120

no i choose a couple: so there are 10 ways to pick any person and 1 way to pick a match and 8 ways to pick any third person

so 10*1*8 = 80

120-80 =40 :?
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If a committee of 3 people is to be selected from among 5 married coup 18 Jul 2021, 21:51
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AbeinOhio
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?

You have to select a committee of 3 people. It means A, B, C form the same committee as B, C, A i.e. you don't have to arrange people. But when you say 'first person' can be selected in 10 ways, you are arranging the 3 people in first, second and third spots. So according to you, A, B, C and B, C, A are different committees.
So all you need to do in un-arrange. To arrange 3 people, you multiply by 3!
To un-arrange, you will divide by 3!
480/3! = 80

There's your answer.

Check out this post for a detailed explanation:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/11 ... nstraints/


VeritasKarishma any idea why my approach didnt work


total number of ways: 5C3 = 120

no i choose a couple: so there are 10 ways to pick any person and 1 way to pick a match and 8 ways to pick any third person

so 10*1*8 = 80

120-80 =40 :?
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You can pick 3 people out of 10 in 10C3 = 120 ways. No problem.

Now, we are not allowed to pick a couple among these 3 people. So cases in which we have a couple are not allowed.
How many such cases are there? Pick one couple out of 5 couples in 5C1 = 5 ways. Pick the third person in 8 ways. So combinations with a couple = 5*8 = 40

Acceptable combinations = 120 - 40 = 80


dave13 : Note that you cannot pick a couple in 10 ways. You are double counting each couple then. Say you pick HubbyA and then pick WifeA is one case. If you pick WifeA and then HubbyA, you are counting this as another case but it is not.
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If a committee of 3 people is to be selected from among 5 married coup 19 Jul 2021, 02:17
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VeritasKarishma thanks so if i am double counting using this method 10*1*8 = 80 then i divide by 2! and get 40. is my method correct now ?
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KarishmaB
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If a committee of 3 people is to be selected from among 5 married coup 21 Jul 2021, 00:18
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VeritasKarishma thanks so if i am double counting using this method 10*1*8 = 80 then i divide by 2! and get 40. is my method correct now ?
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Yes, but picking 10 and then dividing by 2! should make logical sense to you.
I know there are 5 couples so I can pick one couple in 5 ways. That is what makes sense to me.
I am finding it hard to explain to myself why I would pick 10 and then divide by 2!.
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If a committee of 3 people is to be selected from among 5 married coup 22 Apr 2022, 20:16
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Hey KarishmaB


I understand the approach where we are doing (10 * 8 * 6)/ 3! = 80 but I don't understand the other formulaic approach used by some experts including you.

Quote:

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
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I don't understand why we have to do the first step of selecting a couple first that is 5C3

Why can't we just go about directly selecting individuals from the group of 10 people?
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If a committee of 3 people is to be selected from among 5 married coup 23 Apr 2022, 05:36
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Vegita
Hey KarishmaB


I understand the approach where we are doing (10 * 8 * 6)/ 3! = 80 but I don't understand the other formulaic approach used by some experts including you.

Quote:

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

I don't understand why we have to do the first step of selecting a couple first that is 5C3

Why can't we just go about directly selecting individuals from the group of 10 people?
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The individuals are say H1, W1, H2, W2, H3, W3 etc. Now think, how will you ensure that if H1 is selected, W1 is not? If W2 is selected, H2 is not etc.
So easier is to select 3 of the 5 couples and then select one partner from each of those 3 couples. That ways we will certainly not have both partners.
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