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An engagement team consists of a project manager, team

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An engagement team consists of a project manager, team [#permalink] New post 31 Jan 2008, 22:09
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
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Re: combo with constraints [#permalink] New post 01 Feb 2008, 10:42
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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.
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Re: combo with constraints [#permalink] New post 02 Feb 2008, 06:20
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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
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Re: combo with constraints [#permalink] New post 02 Feb 2008, 12:20
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2C1*3C1*7C4 - 2C1*3C1*5C2 = 6[35-10] = 150.
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Re: combo with constraints [#permalink] New post 31 Jan 2008, 22:47
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

So 4+3+2+1 = 10

SO now its 2*3*10 =60

So 210-60 = 150
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Re: combo with constraints [#permalink] New post 01 Feb 2008, 10:15
Following you up till the constraint part...Looking for a combo master to teach us how to do it by calculation and not by hand
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Re: combo with constraints [#permalink] New post 01 Feb 2008, 10:25
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?

25
35
150
210
300


2C1*3C1*(5C4+5C3*2C1)=2*3*(5+10*2)=6*25=150 -> C
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Re: combo with constraints [#permalink] New post 01 Feb 2008, 10:27
Can you explain this part is words:

(5C4+5C3*2C1)
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Re: combo with constraints [#permalink] New post 01 Feb 2008, 10:35
jimmyjamesdonkey wrote:
Can you explain this part is words:

(5C4+5C3*2C1)

yes, can you explain?
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Re: combo with constraints [#permalink] New post 01 Feb 2008, 10:55
Very cool. Thanks. +1
Re: combo with constraints   [#permalink] 01 Feb 2008, 10:55
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