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# A committee of three students has to be formed. There are

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A committee of three students has to be formed. There are [#permalink]

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16 Nov 2010, 23:21
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A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

A. 3
B. 4
C. 5
D. 6
E. 8

m10 q25
[Reveal] Spoiler: OA

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Last edited by Bunuel on 29 May 2014, 00:14, edited 2 times in total.
Renamed the topic and edited the question.
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Re: A committee of three students [#permalink]

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29 Jan 2013, 11:22
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senior wrote:
I'm almost 16 years out of school, I found this one stumped me for a bit.
My preferred approach is to take all possibilities into account, and then backing out the restrictions.

Let's try a different problem with your approach. There are five people: A, B, C, D, E. Need to chose 3 for a committee. A and B cannot be chosen together. B and C cannot be chosen together. How many options?

Your approach: total 10 options, 5c3.
Now, assume the wrong scenario where A and B are chosen together. There are three such scenarios. (A and B are chosen, just need one more person.) So we have to subtract the three wrong options. Similarly, there are three wrong scenarios where B and C are chosen together.

This gives us 10-3-3=4 as the answer.
Yet this answer is wrong. There are five possibilities: ACD, ACE, ADE, BDE, CDE.
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Re: A committee of three students [#permalink]

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16 Nov 2010, 23:48
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scheol79 wrote:
A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

3
4
5
6
8

If Paul is in the committee : We must choose 2 out of 3 others (No Stuart) --> 3 ways
If Stuart is in the committee : No Paul & Hence no Jane, so the other 2 must be chose --> 1 way
If neither Paul nor Stuart are in the committee, 3 students left, but since no Paul, no Jane : No commitee possible

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Re: A committee of three students [#permalink]

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17 Nov 2010, 04:32
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scheol79 wrote:
A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

3
4
5
6
8

Or you can look at it this way: There are two ways to make the committee:
Way1: With Paul
Then to get other two members, you can choose out of 3 people (Stuart cannot be chosen) = 3C2 = 3
Way 2: Without Paul
Then Jane will not be in the committee. So three members needed and only three people to choose from.. 1 way

Total ways of forming the committee = 3 + 1 = 4
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Joined: 01 Feb 2011 Posts: 755 Followers: 14 Kudos [?]: 119 [1] , given: 42 Re: A committee of three students [#permalink] ### Show Tags 12 Jun 2011, 08:20 1 This post received KUDOS bblast wrote: karishma-please let me know where i made a mistake : case 1 : stuart is in the commitee- means only joan and jessica qualify as per conditions so 1 way case 2. PAUL is in the commitee: Jane, Joan and Jessica are other options [strike]hence choosing 3 out of 4 = 4C3 = 4[/strike] [strike]hence answer = 1+4 = 5 [/strike] Paul is in the committee which means Stuart cannot be chosen. So we are left with 3 - Jane,Joan & Jessica. As we have already filled 1 seat with Paul, there are 2 seats left. Thus we are left with choosing 2 out of 3 which is 3c2 = 3 (not 4c3). Hope its clear now. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7187 Location: Pune, India Followers: 2171 Kudos [?]: 14039 [1] , given: 222 Re: A committee of three students [#permalink] ### Show Tags 12 Jun 2011, 14:06 1 This post received KUDOS Expert's post bblast wrote: karishma-please let me know where i made a mistake : case 1 : stuart is in the commitee- means only joan and jessica qualify as per conditions so 1 way case 2. PAUL is in the commitee: Jane, Joan and Jessica are other options hence choosing 3 out of 4 = 4C3 = 4 hence answer = 1+4 = 5 As Spidy001 pointed out, your logic is absolutely fine. You messed up in the last step. It should be 'hence choosing 2 out of 3 (out of Jane, Joan and Jessica)'. We have already chosen Paul. 3C2 = 3 hence answer = 1+3 = 4 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: A committee of three students [#permalink]

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29 Jan 2013, 19:25
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SergeyOrshanskiy wrote:
Let's try a different problem with your approach. There are five people: A, B, C, D, E. Need to chose 3 for a committee. A and B cannot be chosen together. B and C cannot be chosen together. How many options?

Your approach: total 10 options, 5c3.
Now, assume the wrong scenario where A and B are chosen together. There are three such scenarios. (A and B are chosen, just need one more person.) So we have to subtract the three wrong options. Similarly, there are three wrong scenarios where B and C are chosen together.

This gives us 10-3-3=4 as the answer.
Yet this answer is wrong. There are five possibilities: ACD, ACE, ADE, BDE, CDE.

I think you've changed the problem slightly. What you've introduced with your new set of restrictions, is a situation where the restrictions are not mutually exclusive. That is to say, you're counting one of your exclusions twice and it would need to be added back in. When you do 5C3 for both sets, you get ABC as a restriction both times. You simply need to add that duplicate back in as your final step.

Last edited by senior on 30 Jan 2013, 03:32, edited 1 time in total.
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Re: A committee of three students [#permalink]

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29 Jan 2013, 20:36
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SergeyOrshanskiy wrote:
Yes, this is the point I was trying to make. Even worse, if you have three restrictions, you would end up using an inclusion-exclusion formula, such as the one you are using to solve problems about overlapping sets...

Yes, you are right. In fact, I have found that sometimes sets concepts are extremely helpful in solving probability questions. I discussed an interesting sets method to solve a probability question in one of my posts sometime back.

@senior: You may like it so here is the link:
http://www.veritasprep.com/blog/2012/01 ... e-couples/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 13 Jun 2013 Posts: 279 Followers: 13 Kudos [?]: 364 [1] , given: 13 Re: A committee of three students has to be formed. There are [#permalink] ### Show Tags 06 Dec 2014, 04:02 1 This post received KUDOS anceer wrote: can any one explain it with A B C D E as A and B cant be together and B and C cant be separate So we can form BCD, BCE and ADE what will be the another one I am missing hi, B and C can be separate. the question sates that if C is selected then B has to be there. but if B is selected than C may or may not be there. so, considering this in mind. the case you're missing is BDE. i hope it helps. Manager Affiliations: The Earth organization, India Joined: 25 Dec 2010 Posts: 193 WE 1: SAP consultant-IT 2 years WE 2: Entrepreneur-family business 2 years Followers: 5 Kudos [?]: 13 [0], given: 12 Re: A committee of three students [#permalink] ### Show Tags 12 Jun 2011, 01:44 karishma-please let me know where i made a mistake : case 1 : stuart is in the commitee- means only joan and jessica qualify as per conditions so 1 way case 2. PAUL is in the commitee: Jane, Joan and Jessica are other options hence choosing 3 out of 4 = 4C3 = 4 hence answer = 1+4 = 5 _________________ Cheers !! Quant 47-Striving for 50 Verbal 34-Striving for 40 Director Joined: 01 Feb 2011 Posts: 755 Followers: 14 Kudos [?]: 119 [0], given: 42 Re: A committee of three students [#permalink] ### Show Tags 12 Jun 2011, 08:29 restriction 1 : if Paul is chosen then Stuart should not be chosen if Stuart is chosen then Paul should not be chosen. restriction 2 : No Paul no Jane combining these two we have case 1 : if Paul is chosen then Stuart should not be chosen 1c1 * 3c2 = 3 case 2 : if Stuart is chosen then Paul should not be chosen. Jane should be also left out . 1c1 *2c2 = 1 total possible arrangements = 3 +1 = 4. Intern Joined: 06 Jan 2013 Posts: 5 Followers: 0 Kudos [?]: 2 [0], given: 2 Re: A committee of three students [#permalink] ### Show Tags 08 Jan 2013, 04:12 I'm almost 16 years out of school, I found this one stumped me for a bit. My preferred approach is to take all possibilities into account, and then backing out the restrictions. So we start with all possibilities: Step (1) We must choose 3 people of 5, easy enough 5c3 = 10 total possibilities Step (2) Looking at the first restriction, Paul and Stuart do not want to be on the committee together. So let's assume the breaking scenario where Paul or Stuart is selected. That gives us 3c2 scenarios, since we've already selected Paul or Stuart and the other is off the list. So 3c2 = 3 possibilities that should be eliminated Step (3) Jane will only be on the committee with Paul. So we take all the scenarios will Jane will refuse. So if Jane is on the committee, and Paul may not be selected we have 3c2 = 3 possibilities should be eliminated Summary 10 original possibilities - 3 Paul/Stuart restriction - 3 Jane/Paul restriction === 4 remaining possibilities = the answer Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7187 Location: Pune, India Followers: 2171 Kudos [?]: 14039 [0], given: 222 Re: A committee of three students [#permalink] ### Show Tags 29 Jan 2013, 19:51 senior wrote: I think you've changed the problem slightly. What you've introduced with your new set of restrictions, is a situation where the restrictions are not mutually exclusive. That is to say, you're counting one of your exclusions twice and it would need to be added back in. When you do 5C3 for both sets, you get ABC as a restriction both times. You simply need to add that duplicate back in as your final step. Yes, you are right in your analysis. I think, the point Sergey was making was that this approach could make you falter if you are not extremely careful (he's totally right!). If there is an overlap in the two sets you are deducting out of the total, you need to take care to add that back. Personally, I like the 'With B/without B' approach for such questions. There is no overlap possible since they are complementary situations. With B - You cannot choose A and C so you must pick D and E. Only one way Without B - You have 4 options and 4C3 ways to select the committee i.e. 4 ways Total 1 + 4 = 5 ways _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: A committee of three students [#permalink]

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29 Jan 2013, 20:22
VeritasPrepKarishma wrote:
I think, the point Sergey was making was that this approach could make you falter if you are not extremely careful...

Yes, this is the point I was trying to make. Even worse, if you have three restrictions, you would end up using an inclusion-exclusion formula, such as the one you are using to solve problems about overlapping sets...
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29 Jan 2013, 21:28
VeritasPrepKarishma wrote:
@senior: You may like it so here is the link:
http://www.veritasprep.com/blog/2012/01 ... e-couples/

I see. This way we do not have to count the arrangements with exactly one couple sitting together.

This was too complicated for me to think about, so I came up with a different solution.
First arrange two couples. There are three equiprobable arrangements: ([]), ([)], ()[].
In the first one there is a couple sitting together, so the stranger can "spoil" this arrangement in only one of five ways.
In the second one no couple is already sitting together, no matter what the stranger does.
In the third one there will be a couple sitting together, no matter what.
Thus we have 1/3*1/5 + 1/3 + 0 = 6/15=2/5.
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Re: A committee of three students has to be formed. There are [#permalink]

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21 Feb 2014, 07:05
I have question here ..

with paul.. i think we shud take 2c2.. because

becase jane and paul will always be in one team because jane refuse to b in comitee without paul..and no staurt can be with them both.. so we have only choice jessica and joan..

correct me if m wrong
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Re: A committee of three students has to be formed. There are [#permalink]

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23 Feb 2014, 22:11
sanjoo wrote:
I have question here ..

with paul.. i think we shud take 2c2.. because

becase jane and paul will always be in one team because jane refuse to b in comitee without paul..and no staurt can be with them both.. so we have only choice jessica and joan..

correct me if m wrong

Jane refuses to be in the committee without Paul but Paul doesn't refuse to be in the committee without Jane. This means that When Paul is in the committee, Jane may or may not be there. She is one of the candidates we will consider but we needn't 'have to have her in the committee'.

When Paul is not in the committee then Jane is not to be considered.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Current Student Joined: 06 Sep 2013 Posts: 2035 Concentration: Finance GMAT 1: 770 Q0 V Followers: 64 Kudos [?]: 605 [0], given: 355 Re: A committee of three students has to be formed. There are [#permalink] ### Show Tags 28 May 2014, 05:00 So just to get this right and move on. Is it incorrect to do 10C3 - 3C2 (Two of them can't be together) - 3C2 (Jane can only be with Paul, so let's pick the cases in which she needs 1 more member, Paul excluded) = 4 I got to the right answer but not sure if the logic is sound because I've never dealt with two parallel restrictions Please advice Cheers J Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7187 Location: Pune, India Followers: 2171 Kudos [?]: 14039 [0], given: 222 Re: A committee of three students has to be formed. There are [#permalink] ### Show Tags 28 May 2014, 19:26 jlgdr wrote: So just to get this right and move on. Is it incorrect to do 10C3 - 3C2 (Two of them can't be together) - 3C2 (Jane can only be with Paul, so let's pick the cases in which she needs 1 more member, Paul excluded) = 4 I got to the right answer but not sure if the logic is sound because I've never dealt with two parallel restrictions Please advice Cheers J What you need to do is this: 5C3 - 3C1 - 3C2 = 4 5C3 (choose any 3 of 5) 3C1 (Ways of forming a committee having both Paul and Stuart. You need to select only 1 more from the 3 leftover people) 3C2 (Ways of forming a committee with Jane but without Paul. Select 2 of the leftover 3 people) Note that 3C1 has Paul definitely and 3C2 does not have Paul definitely. Hence there will be no overlap in these committees. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: A committee of three students has to be formed. There are [#permalink]

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28 May 2014, 19:31
Right on Karishma, I do understand and acknowledge your method to solve is more clear and less tend to error. I was just wondering If my approach was flawed in any way by the potential overlaps

Thanks again!
Cheers
J

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Re: A committee of three students has to be formed. There are   [#permalink] 28 May 2014, 19:31

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