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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] New post 15 Jan 2011, 16:28
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Question Stats:

63% (02:20) correct 37% (01:42) wrong based on 86 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
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Re: [Urgent]Combination Question! [#permalink] New post 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 #}.

Answer: C.

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] New post 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] New post 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.

Answer: C
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Re: [Urgent]Combination Question! [#permalink] New post 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 #}.

Answer: C.

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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Re: [Urgent]Combination Question! [#permalink] New post 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 #}.

Answer: C.

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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: [Urgent]Combination Question!   [#permalink] 08 Jun 2014, 05:49
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