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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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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]

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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]

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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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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]

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New post 08 Jun 2014, 06:01
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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]

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New post 08 Jun 2014, 06:49
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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 #}.

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.
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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

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

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New post 08 Jun 2015, 01:36
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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New post 08 Jun 2015, 22:32
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.

Final Answer:
[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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New post 22 Jul 2015, 16:35
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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Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]

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New post 26 May 2017, 20:23
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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

1. This is a nCr problem
2. The constraint is , two engineers say e1 and e2 cannot serve together on the same committee
3. It is easier to find the opposite of the constraint and subtract from the total
4. Total number of combinations is 8C3= 56
5. Opposite of the constraint is since e1 and e2 serve together, number of ways of selecting an engineer for the remaining 1 position which is 6C1
6.Total number of combinations with constraints is (4)-(5)=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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New post 27 May 2017, 04:51
Case 1. We exclude BOTH of them, so we have to choose 3 from remaining 6
= 6C3 = 20 ways

Case 2. We include the first inexperienced engineer but exclude the other, now we have to choose 2 from remaining 6 engineers (As we cannot consider the other inexperienced one) = 6C2 = 15 ways

Case 3. Include second inexperience engineer, exclude the first one, same as previous = 15 ways

Total ways = 20+15+15 = 50

Hence answer is 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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New post 01 Jun 2017, 09:53
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


We can use the following formula:

Total number of ways to select the committee = # of ways with both inexperienced engineers selected + # of ways with two inexperienced engineers NOT selected

Thus:

# of ways with two inexperienced engineers NOT selected = Total number of ways to select the committee - # of ways with both inexperienced engineers selected

Total number of ways to select the committee:

8C3 = (8 x 7 x 6)/3! = 56 ways

Now let’s calculate the total number of ways to select the committee such that the two inexperienced engineers are both selected. One such occurrence would be:

NNE (N = inexperienced engineer and E = experienced engineer)

Since both inexperienced engineers have been selected, there is only 1 position left, and there are 6 experienced engineers to fill it. Thus, we have:

6C1 = 6

Thus, the number of ways to select the committee members with both inexperienced engineers NOT selected = 56 - 6 = 50.

Answer: 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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New post 13 Jun 2017, 12:41
The 3 member committee can be made in 2 ways:
Case 1: E E I=6C2*2C1=30
Case 2: E E E=6C3=20
Thus total ways 30+20=50

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Re: Ben needs to form a committee of 3 from a group of 8 enginee   [#permalink] 13 Jun 2017, 12:41
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