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crackgmat007
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IMO A
Paul,Jane,Joan
Paul,Jane,Jessica
Stuart,Joan,Jessica
Committee
This topic has expert replies
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tohellandback
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vivekjaiswal
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tohellandback,tohellandback wrote:IMO A
Paul,Jane,Joan
Paul,Jane,Jessica
Stuart,Joan,Jessica
I think the combination Paul, Joan, Jessica has been missed out.
Its said that Jane won't go without Paul, but Paul can be selected without keeping Jane in the committee.
Thus the total number of possibilities are 4, hence answer should be B.
Is there a better way to do this?
cheers,
Vivek
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blr_gmat_prep
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Selecting 5 people out of 3 people = 5c3=10
Removing combinations where Paul and Stuart are together. P, S are fixed. Remaining one place can be filled by any of the 3 people so total ways => 3
Removing combincations where Paul and Jessica are not together. If either Paul or jessica are in committee and other is not there, so remaining place can be filled by 3c2 people => 3
So Total => 10-3-3 = 4
Removing combinations where Paul and Stuart are together. P, S are fixed. Remaining one place can be filled by any of the 3 people so total ways => 3
Removing combincations where Paul and Jessica are not together. If either Paul or jessica are in committee and other is not there, so remaining place can be filled by 3c2 people => 3
So Total => 10-3-3 = 4
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Stuart@KaplanGMAT
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First, I'd like to say that Paul is a jerk and that I can't believe that Jane took his side!crackgmat007 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
OA - B
Second, while we could use formulas, a lot of combination questions are easier to answer through a bit of common sense and brute force. On this question, it's much quicker to just list the possible committees than it is to use a formula approach.
No Paul means no Jane (that traitor!), which leaves us with only Joan/Stuart/Jessica. Sadly, I cannot appear in any other committees.
Paul then needs 2 co-members and has 3 possible buddies for those 2 spots. We could use 3C2=3 to calculate, or we could just brute force:
Paul/Joan/Jessica
Paul/Joan/Jane
Paul/Jane/Jessica
So, as others have noted, there are 4 possible committees - choose (B).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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beatthegmat
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Hahaha!Stuart Kovinsky wrote:First, I'd like to say that Paul is a jerk and that I can't believe that Jane took his side!
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