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

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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.

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

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

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

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Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]
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]
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 $$C^2_2*C^1_6=6$$ 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: https://gmatclub.com/forum/math-combinatorics-87345.html

Hope it helps.

Thanks for the solution. For the second half that we are subtracting - is the reason why we are not doing 2*6 because positions dont matter here?
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Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]
kanikab 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 $$C^2_2*C^1_6=6$$ 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: https://gmatclub.com/forum/math-combinatorics-87345.html

Hope it helps.

Thanks for the solution. For the second half that we are subtracting - is the reason why we are not doing 2*6 because positions dont matter here?

When counting the number of committees with two inexperienced engineers, the number of ways to choose any but those two is indeed 6 (or as it's written in my solution 6C1, which is 6). But the number of ways to choose the two inexperienced engineers from two, is not 2, it's 1 (only 1 way to choose two out of two). So, you get 1*6 (or as it's written in my solution 2C2*6C1, which is 1*6).
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Re: Ben needs to form a committee of 3 from a group of 8 enginee [#permalink]
Bunuel wrote:
kanikab 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 $$C^2_2*C^1_6=6$$ 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: https://gmatclub.com/forum/math-combinatorics-87345.html

Hope it helps.

Thanks for the solution. For the second half that we are subtracting - is the reason why we are not doing 2*6 because positions dont matter here?

When counting the number of committees with two inexperienced engineers, the number of ways to choose any but those two is indeed 6 (or as it's written in my solution 6C1, which is 6). But the number of ways to choose the two inexperienced engineers from two, is not 2, it's 1 (only 1 way to choose two out of two). So, you get 1*6 (or as it's written in my solution 2C2*6C1, which is 1*6).

Got it, thank you! My bad, I kept writing nCn (2C2) as nC1(2C1) even though I was thinking right in my head hence the confusion.
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Re: Nicole needs to form a committee of 3 from a group of 8 research attor [#permalink]
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Re: Nicole needs to form a committee of 3 from a group of 8 research attor [#permalink]
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