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 500,000 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:

Consider the number of way a committee of 3 can be selected [#permalink]

Show Tags

20 Apr 2010, 12:09

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

100% (01:50) correct
0% (00:00) wrong based on 8 sessions

HideShow timer Statistics

Consider the number of way a committee of 3 can be selected from 7 people- A, B, C, D, E, F, G if order does not matter, and C and E cannot be chosen together.

Consider the number of way a committee of 3 can be selected from 7 people- A,B,C,D,E,F,G if:

Order does not matter,

and C and E cannot be chosen together.

What is the justification?

\(C^3_7-C^1_5=30\)

\(C^3_7\) - # of ways we can choose any 3 out of 7 (without restriction). \(C^1_5\) - # of groups with C and E together (if C and E are in chosen group, then third member can be any out of 5 left, so total # of groups is \(C^1_5\)).
_________________

Because only 1 way to choose C and 1 Way to choose E?

Therefore you minus that from the complete total ways to pick 3 from 7?

Yes. You can write this as \(C^1_1*C^1_1*C^1_5=5\) (one way to choose C and one way to choose choose E) OR \(C^2_2*C^1^5=5\) (one way to choose C and E, from C and E), which is basically \(C^1_5=5\).

ksharma12 wrote:

Essentially the same thing as at least 1 problem?

If you mean in a way: total-opposite, then yes.
_________________

Re: combination problem justification [#permalink]

Show Tags

10 Sep 2012, 00:11

Bunuel wrote:

ksharma12 wrote:

Consider the number of way a committee of 3 can be selected from 7 people- A,B,C,D,E,F,G if:

Order does not matter,

and C and E cannot be chosen together.

What is the justification?

\(C^3_7-C^1_5=30\)

\(C^3_7\) - # of ways we can choose any 3 out of 7 (without restriction). \(C^1_5\) - # of groups with C and E together (if C and E are in chosen group, then third member can be any out of 5 left, so total # of groups is \(C^1_5\)).

just for discussion sake and to clear the concept I was wondering what would happen if the order mattered in the above scenario. Eva ..Bunuel and others in the community request you to please contribute .

lets assume ABCDEFG is a 7 letter word so ABC is different from CAB meaning order matters so if I wanted to form 3 letter words such that CE should not be together then , what would be the number of ways this could be done ?

I tried to a certain extent please verify if this is correct or not.

\(C^7_3 * 3! - C^1_1 *C^1_1*C^5_1*3!\)=180

\(C^7_3\) = # of ways to select 3 out of 7 3! = # of ways to arrange them among themselves as order matters

\(C^1_1\)= choosing c \(C^1_1\) = choosing e \(C^5_1\)= # of ways to select one from remaining 5 , after C and E have been chosen in the community. \(3!\) = # of to arrange the 3 letter committee containing both C and E

Re: combination problem justification [#permalink]

Show Tags

10 Sep 2012, 04:11

1

This post received KUDOS

stne wrote:

Bunuel wrote:

ksharma12 wrote:

Consider the number of way a committee of 3 can be selected from 7 people- A,B,C,D,E,F,G if:

Order does not matter,

and C and E cannot be chosen together.

What is the justification?

\(C^3_7-C^1_5=30\)

\(C^3_7\) - # of ways we can choose any 3 out of 7 (without restriction). \(C^1_5\) - # of groups with C and E together (if C and E are in chosen group, then third member can be any out of 5 left, so total # of groups is \(C^1_5\)).

just for discussion sake and to clear the concept I was wondering what would happen if the order mattered in the above scenario. Eva ..Bunuel and others in the community request you to please contribute .

lets assume ABCDEFG is a 7 letter word so ABC is different from CAB meaning order matters so if I wanted to form 3 letter words such that CE should not be together then , what would be the number of ways this could be done ?

I tried to a certain extent please verify if this is correct or not.

\(C^7_3 * 3! - C^1_1 *C^1_1*C^5_1*3!\)=180

\(C^7_3\) = # of ways to select 3 out of 7 3! = # of ways to arrange them among themselves as order matters

\(C^1_1\)= choosing c \(C^1_1\) = choosing e \(C^5_1\)= # of ways to select one from remaining 5 , after C and E have been chosen in the community. \(3!\) = # of to arrange the 3 letter committee containing both C and E

Hope this is correct?

Yes, this is correct.

I just have my own preferences to count...so, for example, choose 3 out of 7 when order matters I write directly 7*6*5, meaning I directly take into account the order (why write the formula with the factorials for nCk and then multiply by k!, reduce...). First choice 7 options, second choice 6, third 5. And I am not even writing the 1C1 factors. There is nothing wrong with it, but I know that C and E must be chosen, then I need just one extra person (letter ), so I can choose 1 out of 5, for which again I am not writing the 5C1, and then having C, E and * (somebody), I have to consider all the permutations of the three, so 3!...
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Consider the number of way a committee of 3 can be selected [#permalink]

Show Tags

22 Jan 2014, 16:31

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.
_________________

Re: Consider the number of way a committee of 3 can be selected [#permalink]

Show Tags

22 Jan 2014, 17:26

Bunuel, can we also think of it as 6C3 + 6C3 - 3C5?

I am trying to calculate combinations each C and E with other 5 separately, adding them together, then subtracting out one of the double counted 3 team combinations of the other 5 members. Not sure if this is a proper approach? Thanks!
_________________

Bunuel, can we also think of it as 6C3 + 6C3 - 3C5?

I am trying to calculate combinations each C and E with other 5 separately, adding them together, then subtracting out one of the double counted 3 team combinations of the other 5 members. Not sure if this is a proper approach? Thanks!

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...