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Ben needs to form a committee of 3 from a group of 8 enginee

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

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15 Jan 2011, 16:28
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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
[Reveal] Spoiler: OA
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15 Jan 2011, 16:46
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eltonli wrote:
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 #}.

Check Combinatoric chapter of Math Book for more: math-combinatorics-87345.html

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

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28 May 2014, 11:52
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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.
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Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]

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04 Jun 2014, 05:46
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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.

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08 Jun 2014, 05:01
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Bunuel wrote:
eltonli wrote:
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 #}.

Check Combinatoric chapter of Math Book for more: math-combinatorics-87345.html

Hope it helps.

Hi Bunuel,
when you write: $$C^2_2*C^2_6=6$$ you actually mean this: $$C^2_2*C^1_6=6$$ right?
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08 Jun 2014, 05:49
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ronr34 wrote:
Bunuel wrote:
eltonli wrote:
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 #}.

Check Combinatoric chapter of Math Book for more: math-combinatorics-87345.html

Hope it helps.

Hi Bunuel,
when you write: $$C^2_2*C^2_6=6$$ you actually mean this: $$C^2_2*C^1_6=6$$ right?

Right. Edited the typo. Thank you.
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Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]

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08 Jun 2015, 00:36
good question.
how many choices we can have if two unexperienced engineers can work togetther

after choosing two unexperienced engineer , we take another engineer from 6 remaining. totally we have 6 combination.

from 8 engineers, we can have 8.7.6/3.2.1=56 combination

we need to minus
50

princeton review give us explanation of counting and combination . we should read this part.
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Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]

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08 Jun 2015, 21:32
Hi All,

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.

15+15+10 = 50 total options.

[Reveal] Spoiler:
C

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

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22 Jul 2015, 15:35
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
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Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]

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11 Sep 2016, 02:49
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Re: Ben needs to form a committee of 3 from a group of 8 enginee   [#permalink] 11 Sep 2016, 02:49
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