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An engagement team consists of a project manager, team leade [#permalink]
31 Jan 2008, 21:09

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00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

67% (03:02) correct
33% (02:18) wrong based on 213 sessions

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?

Re: combo with constraints [#permalink]
31 Jan 2008, 21:47

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marcodonzelli wrote:

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

Re: combo with constraints [#permalink]
01 Feb 2008, 09:25

marcodonzelli wrote:

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?

Re: combo with constraints [#permalink]
01 Feb 2008, 09:42

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jimmyjamesdonkey wrote:

Can you explain this part is words:

(5C4+5C3*2C1)

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.

Re: combo with constraints [#permalink]
02 Feb 2008, 05:20

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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]
18 Feb 2014, 03:33

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Re: An engagement team consists of a project manager, team leade [#permalink]
14 May 2014, 02:46

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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]
17 May 2014, 01:25

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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).

gmatclubot

Re: An engagement team consists of a project manager, team leade
[#permalink]
17 May 2014, 01:25

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