Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A group of 10 people consists of 3 married couples and 4 [#permalink]
17 May 2011, 07:18

1

This post received KUDOS

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
100% (00:49) wrong based on 7 sessions

A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method --------------- No of ways to select one couple and two singles + No of ways to select all singles

Re: A group of 10 people [#permalink]
17 May 2011, 08:27

Complicated method -

You have to consider all the possibilities.

Obvious cases - All 4 singles & 1 married couple and 2 singles

Other cases - 1 marrried person from a couple and 3 singles 1 married couple, 1 married person from a couple and 1 single 1 married couple, 2 married person from different couples 1 married person from each couple and 1 single 1 married person from 2 separate couples and 2 singles

Re: A group of 10 people [#permalink]
17 May 2011, 20:50

1

This post was BOOKMARKED

gmat1220 wrote:

A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method --------------- No of ways to select one couple and two singles + No of ways to select all singles

Let A----B C----D E ----F are couples and 1 2 3 4 are singles

1st scenario .. 4 singles out of 4 singles ------- 4c4 = 1 2nd scenario 2 Married singles + 2 singles ------ ( 6*4)/2 * 4c2 = 72 3rd 1 Married single + 3 singles ------- 6*4c3 = 24 4th 3 married single + 1 single ------- (6*4*2)/3!*4C1 = 32 5th scenario 1Couple + 1 Married single + 1 single ---- 3c1*4*4c1= 48 6th scenario 1 couple + 2 Married singles ------- 3c1*( 4*2)/2 = 12 7th scenario 1 couple + 2 singles --------- 3c1*4C2 = 18

Re: A group of 10 people [#permalink]
14 Sep 2011, 12:24

sudhir18n wrote:

gmat1220 wrote:

A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method --------------- No of ways to select one couple and two singles + No of ways to select all singles

Let A----B C----D E ----F are couples and 1 2 3 4 are singles

1st scenario .. 4 singles out of 4 singles ------- 4c4 = 1 2nd scenario 2 Married singles + 2 singles ------ ( 6*4)/2 * 4c2 = 72 3rd 1 Married single + 3 singles ------- 6*4c3 = 24 4th 3 married single + 1 single ------- (6*4*2)/3!*4C1 = 32 5th scenario 1Couple + 1 Married single + 1 single ---- 3c1*4*4c1= 48 6th scenario 1 couple + 2 Married singles ------- 3c1*( 4*2)/2 = 12 7th scenario 1 couple + 2 singles --------- 3c1*4C2 = 18

Total = 1+72+24+ 32+48+12+18 = 207

hi - you could have simply done this ..

Total number of ways of selecting 4 people out of 10 = 10C4 Total number of ways of selecting 2 married couple = 3C2

Re: A group of 10 people [#permalink]
10 Oct 2011, 05:32

total number of combinations 10c4 = 10!/4!6! = 210 total number of married couple = 3c2 = 3!/2! = 3 total number of possible commities are 207 _________________

Re: A group of 10 people [#permalink]
16 Nov 2011, 15:38

We can use the other way of thinking, that is (total possibility of selecting 4 people)-(selecting two couples) then the answer will be 10C4-3C2=210-3=207. That's much more quickly

Re: A group of 10 people [#permalink]
13 Feb 2013, 17:27

One way to do problem: Answer = No of ways to select one couple and two singles + No of ways to select all singles=207=official answer

How about the 1-x method

"The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207"

Since when has the opposite of "at most 1" equaled to exact no of ways to select 2 couples? Could someone help me grasp why this is conceptually the opposite of exactly 1 couple? _________________

Re: A group of 10 people [#permalink]
13 Feb 2013, 19:31

Expert's post

1

This post was BOOKMARKED

manimgoindowndown wrote:

One way to do problem: Answer = No of ways to select one couple and two singles + No of ways to select all singles=207=official answer

How about the 1-x method

"The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207"

Since when has the opposite of "at most 1" equaled to exact no of ways to select 2 couples? Could someone help me grasp why this is conceptually the opposite of exactly 1 couple?

You need to select a committee of 4. You can do it in 3 ways: a. Select no couple and 4 others b. Select one couple and 2 others c. Select two couples

'At most one' means either no couples or only one couple. What is the complement of at most one? The only other way in which you can select a committee of 4 is by selecting 2 couples. So you can find the number of ways of selecting 2 couples and subtract that from the total number of ways of selecting 4 people. You will get the number of ways in which there will be at most one couple. _________________

Re: A group of 10 people [#permalink]
15 Feb 2013, 03:47

VeritasPrepKarishma wrote:

manimgoindowndown wrote:

One way to do problem: Answer = No of ways to select one couple and two singles + No of ways to select all singles=207=official answer

How about the 1-x method

"The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207"

Since when has the opposite of "at most 1" equaled to exact no of ways to select 2 couples? Could someone help me grasp why this is conceptually the opposite of exactly 1 couple?

You need to select a committee of 4. You can do it in 3 ways: a. Select no couple and 4 others b. Select one couple and 2 others c. Select two couples

'At most one' means either no couples or only one couple. What is the complement of at most one? The only other way in which you can select a committee of 4 is by selecting 2 couples. So you can find the number of ways of selecting 2 couples and subtract that from the total number of ways of selecting 4 people. You will get the number of ways in which there will be at most one couple.

Namaste aur shukria so much again Karishma. Made it clear within the context of the problem _________________

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]
16 Feb 2013, 03:10

Expert's post

gmat1220 wrote:

A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method --------------- No of ways to select one couple and two singles + No of ways to select all singles

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]
22 Feb 2013, 06:47

my question is : i understand we can do like selecting 2 couples and subtract that from the total number of ways of selecting 4 people. now there 6 married peple and 4 singles so it should be 10c4 - 6x5x4x3/4x3x2x1

but i dont understand how 10c4- 3x2/2 comes?? can someone please explain. _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]
22 Feb 2013, 07:04

1

This post received KUDOS

Expert's post

skamal7 wrote:

my question is : i understand we can do like selecting 2 couples and subtract that from the total number of ways of selecting 4 people. now there 6 married peple and 4 singles so it should be 10c4 - 6x5x4x3/4x3x2x1

but i dont understand how 10c4- 3x2/2 comes?? can someone please explain.

We have 3 couples, so 3C2=3!/2!=3 is the number of ways to choose 2 couples out of 3.

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]
22 Feb 2013, 07:10

bunnel thanks for kind response.

One thing i dont understand is while taking total number of ways of selecting 4 people we count the 3 married couples as 6 right.?so when we are selecting 2 married couples it should be 6x5x4x3/4! na??but why when selecting 2 married couples we are taking count just as 3x2/2!?? please do explain!!! _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]
22 Feb 2013, 07:13

1

This post received KUDOS

Expert's post

skamal7 wrote:

bunnel thanks for kind response.

One thing i dont understand is while taking total number of ways of selecting 4 people we count the 3 married couples as 6 right.?so when we are selecting 2 married couples it should be 6x5x4x3/4! na??but why when selecting 2 married couples we are taking count just as 3x2/2!?? please do explain!!!

We are interested in number of ways we can select 2 couples out of 3, so 2 entities out of 3. _________________

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]
13 Aug 2014, 07:23

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...