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:

Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]

Show Tags

15 Jan 2011, 17:28

2

This post received KUDOS

13

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

60% (01:12) correct
40% (01:23) wrong based on 374 sessions

HideShow timer Statistics

Ben needs to form a committee of 3 from a group of 8 engineers. If two of the engineers are too inexperienced to serve together on the committee, how many different committees can Ben form?

Have no idea how to approach this. Appreciate if you guys can help out -Ben needs to form a committee of 3 from a group of 8 engineers. If two of the engineers are too inexperienced to serve together on the committee, how many different committees can Ben form? A.20 B.30 C.50 D.56 E.336

\(C^3_8-C^2_2*C^1_6=56-6=50\): where \(C^3_8=56\) is total ways to choose 3 engineers out of 8 and \(C^2_2*C^1_6=6\) is ways to choose those 2 who are inexperienced with any from 6 remaining for the third member, so you get {total committees possible} - {committees you don't want to form} = {desired #}.

Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]

Show Tags

28 May 2014, 12:52

6

This post received KUDOS

1

This post was BOOKMARKED

6-E || 2-I

Case 1: EEI = 6C2 * 2C1 = 30 Case 2: EEE = 6C3 = 20

EII (not allowed as per the question)

Thus (C) 20 + 30 = 50 Answer.
_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]

Show Tags

04 Jun 2014, 06:46

1

This post received KUDOS

eltonli wrote:

Ben needs to form a committee of 3 from a group of 8 engineers. If two of the engineers are too inexperienced to serve together on the committee, how many different committees can Ben form?

A. 20 B. 30 C. 50 D. 56 E. 336

Since 8 choose 3 is 56, we know the answer has to be A, B, or C. That narrows down our choices a lot already.

Now, imagine our 8 engineers look like this:

| | | | | | | |

where the first two are the ones who are too inexperienced to work together. If we have these two together in a group, then we have 6 remaining options for the last member of the group. Thus, there are only 6 possible committees with these two together.

56 total possibilities - 6 groups with these two together = 50 committees.

Have no idea how to approach this. Appreciate if you guys can help out -Ben needs to form a committee of 3 from a group of 8 engineers. If two of the engineers are too inexperienced to serve together on the committee, how many different committees can Ben form? A.20 B.30 C.50 D.56 E.336

\(C^3_8-C^2_2*C^1_6=56-6=50\): where \(C^3_8=56\) is total ways to choose 3 engineers out of 8 and \(C^2_2*C^2_6=6\) is ways to choose those 2 who are inexperienced with any from 6 remaining for the third member, so you get {total committees possible} - {committees you don't want to form} = {desired #}.

Have no idea how to approach this. Appreciate if you guys can help out -Ben needs to form a committee of 3 from a group of 8 engineers. If two of the engineers are too inexperienced to serve together on the committee, how many different committees can Ben form? A.20 B.30 C.50 D.56 E.336

\(C^3_8-C^2_2*C^1_6=56-6=50\): where \(C^3_8=56\) is total ways to choose 3 engineers out of 8 and \(C^2_2*C^2_6=6\) is ways to choose those 2 who are inexperienced with any from 6 remaining for the third member, so you get {total committees possible} - {committees you don't want to form} = {desired #}.

There are a couple of different ways to approach the 'math' in this question. Here's another way that breaks the possibilities into smaller groups:

We're told t0 form a committee of 3 from a total of 8 engineers. HOWEVER, two of the engineers are too inexperienced to be on the committee together, so we cannot allow that...

Let's call the two inexperienced engineers A and B; then there are the other 6 engineers who DO have enough experience.

There are now 3 options when it comes to forming a committee: 1) Engineer "A" and 2 experienced engineers. 2) Engineer "B" and 2 experienced engineers. 3) 3 experienced engineers

For the first option, we can choose 2 engineers from the 6 experienced engineers = 6C2 = 6!/(2!4!) = 15 options. For the second option, we can choose 2 engineers from the 6 experienced engineers = 6C2 = 6!/(2!4!) = 15 options. For the third option, we can choose 3 engineers from the 6 experienced engineers = 6C3 = 6!/(3!3!) = 10 options.

Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]

Show Tags

22 Jul 2015, 16:35

Total Number of Committees possible = 8C3 = 56 Total Number of Committees were the two inexperienced say A,B can be together AB_ = 6C1 or 6 Subtracting 56-6 = 50

Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]

Show Tags

11 Sep 2016, 03:49

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.
_________________

Ben needs to form a committee of 3 from a group of 8 engineers. If two of the engineers are too inexperienced to serve together on the committee, how many different committees can Ben form?

A. 20 B. 30 C. 50 D. 56 E. 336

1. This is a nCr problem 2. The constraint is , two engineers say e1 and e2 cannot serve together on the same committee 3. It is easier to find the opposite of the constraint and subtract from the total 4. Total number of combinations is 8C3= 56 5. Opposite of the constraint is since e1 and e2 serve together, number of ways of selecting an engineer for the remaining 1 position which is 6C1 6.Total number of combinations with constraints is (4)-(5)=50
_________________

Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]

Show Tags

27 May 2017, 04:51

Case 1. We exclude BOTH of them, so we have to choose 3 from remaining 6 = 6C3 = 20 ways

Case 2. We include the first inexperienced engineer but exclude the other, now we have to choose 2 from remaining 6 engineers (As we cannot consider the other inexperienced one) = 6C2 = 15 ways

Case 3. Include second inexperience engineer, exclude the first one, same as previous = 15 ways

Ben needs to form a committee of 3 from a group of 8 engineers. If two of the engineers are too inexperienced to serve together on the committee, how many different committees can Ben form?

A. 20 B. 30 C. 50 D. 56 E. 336

We can use the following formula:

Total number of ways to select the committee = # of ways with both inexperienced engineers selected + # of ways with two inexperienced engineers NOT selected

Thus:

# of ways with two inexperienced engineers NOT selected = Total number of ways to select the committee - # of ways with both inexperienced engineers selected

Total number of ways to select the committee:

8C3 = (8 x 7 x 6)/3! = 56 ways

Now let’s calculate the total number of ways to select the committee such that the two inexperienced engineers are both selected. One such occurrence would be:

NNE (N = inexperienced engineer and E = experienced engineer)

Since both inexperienced engineers have been selected, there is only 1 position left, and there are 6 experienced engineers to fill it. Thus, we have:

6C1 = 6

Thus, the number of ways to select the committee members with both inexperienced engineers NOT selected = 56 - 6 = 50.

Answer: C
_________________

Scott Woodbury-Stewart Founder and CEO

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

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...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...

“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...