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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, 17:28

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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?

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 #}.

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

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28 May 2014, 12:52

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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, 06:46

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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.

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 #}.

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 #}.

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.

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

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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

Re: 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

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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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
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27 May 2017, 04:51

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