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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]  15 Jan 2011, 16:28
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61% (02:21) correct 39% (01:42) wrong based on 97 sessions
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
Math Expert
Joined: 02 Sep 2009
Posts: 27056
Followers: 4184

Kudos [?]: 40371 [1] , given: 5420

Re: [Urgent]Combination Question! [#permalink]  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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Status: After a long voyage, land is visible.
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Location: India
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Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]  28 May 2014, 11:52
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KUDOS
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]  04 Jun 2014, 05:46
1
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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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Re: [Urgent]Combination Question! [#permalink]  08 Jun 2014, 05:01
1
KUDOS
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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Posts: 27056
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Re: [Urgent]Combination Question! [#permalink]  08 Jun 2014, 05:49
Expert's post
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: [Urgent]Combination Question!   [#permalink] 08 Jun 2014, 05:49
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