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:

An engagement team consists of a project manager, team leade [#permalink]

Show Tags

31 Jan 2008, 21:09

3

This post received KUDOS

27

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

68% (02:58) correct
32% (02:06) wrong based on 651 sessions

HideShow timer Statistics

An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?

An engagement team consists of a project manager, team leader and four consultants. There are 2 candidates for the position of proj.manager, 3 for the position of team leader and 7 for 4 consultants slots. If 2 of the 7 consultants refuse to be on the same team, how many different team are possible?

25 35 150 210 300

I get C. If I see this on the test, id prolly get to 210 and then guess A B or C.

OK so for the first position we have only 2 possiblities ( proj manager) 3 for the team leader so and 7!/3!4! for the consultants

2*3*35 --> 210

Now I dunno how to figure out the constraints quickly but I eventually figured it out. Obvs. we want to figure out the total ways in which the two ARE on the same team.

I did it by AB are the two and XYZFN are the rest

ABXO (O stands for the other 4) So the first is ABXY, Z,F,N 4 possible choices

The next is

ABYO (but notice we already had YX so there are only 3 possible choices) ABZO ABFO ABNO No possible here we used them all up

An engagement team consists of a project manager, team leader and four consultants. There are 2 candidates for the position of proj.manager, 3 for the position of team leader and 7 for 4 consultants slots. If 2 of the 7 consultants refuse to be on the same team, how many different team are possible?

We can either choose 4 consultants from a group of 5 (5C4) who are willing to work with everyone or choose 3 from that group (5C3) and one person from the other group of 2 consultants (2C1) who don't want to work with each other.

To get the number of ways you can select 4 consultants, you can do this:

A. Total number of ways 4 can be selected out of 7 = 7C4 = 35 B. Total number of ways in which these two snobbish consultants are together (two are already there, so you just need to select rest two out of 5) = 5C2 = 10

A (minus) B results in 25 which can then be multiplied with (2C1*3C1) to result in 150

Re: An engagement team consists of a project manager, team [#permalink]

Show Tags

18 Feb 2014, 03:33

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: An engagement team consists of a project manager, team leade [#permalink]

Show Tags

14 May 2014, 02:46

2

This post received KUDOS

2

This post was BOOKMARKED

Answer: C a) No of ways to select 1 Manager = 2c1 = 2 b) No of ways to select 1 Team leader = 3c1 = 3 c) No of ways to select 4 Consultants = 7c4 = 35 Therefore, possible teams without any constraint = 2x3x35 = 210

No of ways to select 4 Consultants out of 7 when 2 of them are always together = 6c4 x2! = 60

Therefore, possible teams with given constraint = 210 - 60 = 150

Re: An engagement team consists of a project manager, team leade [#permalink]

Show Tags

17 May 2014, 01:25

5

This post received KUDOS

Well, probably the quickest way to do this problem is to

1) eliminate D & E, because we are pretty sure it is less than 210 (we know max is 2*3*35). 2) eliminate A & B, because we know 25 and 35 are not multiple of 6 (we know the combination will be an integer).

Re: An engagement team consists of a project manager, team leade [#permalink]

Show Tags

29 Jun 2015, 10:24

2

This post received KUDOS

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.
_________________

An engagement team consists of a project manager, team leade [#permalink]

Show Tags

21 Nov 2015, 11:21

1

This post received KUDOS

srp wrote:

To get the number of ways you can select 4 consultants, you can do this:

A. Total number of ways 4 can be selected out of 7 = 7C4 = 35 B. Total number of ways in which these two snobbish consultants are together (two are already there, so you just need to select rest two out of 5) = 5C2 = 10

A (minus) B results in 25 which can then be multiplied with (2C1*3C1) to result in 150

Simple and straightforward! (Also, I used this method as well! )

I'd like to add that your B is actually 2C2 * 5C2 = 1 * 10 = 10 We have 2C2, because we have already chosen the consultants in the group.

Re: An engagement team consists of a project manager, team leade [#permalink]

Show Tags

13 Jul 2016, 04:05

marcodonzelli wrote:

An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?

A. 25 B. 35 C. 150 D. 210 E. 300

This is how I solved this question and got the answer incorrect.

Total Combinations - (the number of combinations in which A and B are always together) = Number of ways in which A and B won't work together

Assuming that A and B are the two consultants who don't want to work together.

Number of ways to choose 1 PM out of 2 candidates x Number of ways to choose 1 TL out of 3 candidates x Number of ways to choose 4 Consultants out of 7 candidates = Total

2 ways x 3 ways x 35 ways(7!/4!.3!) = 210 ways.

I got this right. However, somewhere in the next steps is where I made a mistake.

Number of ways in which A and B will work together and the rest two positions can be filled with the remaining 5 candidates.

2 x 3 x 1 x 1 x 5 x 4

So I subtracted 120 from 210.

However after following the answers here, I understood that for the remaining two positions, instead of doing a 5C2, I chose to fill them in 5 ways and 4 ways.

I would like to understand: 1) what sort of an error am I committing. 2When should I follow the approach I used above and when should I do a 5C2 logic.

Re: An engagement team consists of a project manager, team leade [#permalink]

Show Tags

13 Jul 2016, 22:42

total teams 2c1 * 3c1* 7c4 = 210 let both be on same team and always selected so now we have to choose 2 out of remaining 5 total team when both candidates on same team will be 2c1 * 3c1* 5c2 = 60 210-60 = 150

The answer choices to this question provide us with an interesting 'shortcut' that we can use to avoid some of the math involved.

Since there are 2 possible project managers and 3 potential team leaders, then the final answer MUST be a multiple of (2)(3) = 6.

Since we're choosing 4 of 7 possible consultants, we can use the Combination Formula:

7!/(4!3!) = 35 possible groups of 4 consultants.

IF there were no additional restrictions, then there would be (6)(35) = 210 possible groups. HOWEVER, we know that certain consultants won't work with other consultants, so the number of possible groups must be LESS than 210. Based on the answer choices, there's only one option that is a multiple of 6 and is less than 210...

Re: An engagement team consists of a project manager, team leade [#permalink]

Show Tags

22 Dec 2016, 00:34

marcodonzelli wrote:

An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?

A. 25 B. 35 C. 150 D. 210 E. 300

i did this question in this way 2c1 * 3c1* 5c4+5c3*2c1= 2*3*(5+5*4*2/2)=6*25=150

gmatclubot

Re: An engagement team consists of a project manager, team leade
[#permalink]
22 Dec 2016, 00:34

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...