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A committee of 3 students has to be formed. The are five [#permalink]
 28 Jul 2007, 12:05
A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3
(ii) 4
(iii) 5
(iv) 6
(v) 8
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Senior Manager
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Re: Tough P&C Question [#permalink]
28 Jul 2007, 12:12
ajay_gmat wrote: A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3
(ii) 4
(iii) 5
(iv) 6
(v) 8
Should be (ii).
If Paul is in the committee there can be 3C2 = 3 committees. (Stuart won't be there.)
I Stuart is in the committee there can only be 1 committee. (Paul and Jane won't be there.)
There can be no committee without either of them. (Stuart, Paul and Jane won't be there, so there are only 2 left.)
So total number of possible committees is 4.
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Senior Manager
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Re: Tough P&C Question [#permalink]
28 Jul 2007, 12:15
ajay_gmat wrote: A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3
(ii) 4
(iii) 5
(iv) 6
(v) 8
there are 5C3 ways to make a comittee of 3 students out of 5 w/o restriction. = 10
there are 3 ways to have a team with Paul and Stuart
there are 3 ways to have a team with Jane and Paul
There is 1 way to have a team with Jane Paul and Stewart, but it is counted twice above.
i say D - 6
edit: i read the question wrong.
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Director
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Re: Tough P&C Question [#permalink]
28 Jul 2007, 12:31
ajay_gmat wrote: A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3
(ii) 4
(iii) 5
(iv) 6
(v) 8
A. i would say 3.
(jean + Paul) + (either joan or jessica) = 2
(paul) + (joan and jessica) (i.e. not stuart and jean) = 1
so total = 3
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Director
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Agree B.
3C1 (Paul and Jane together) +1 (no Paul and Jane) = 4.
Last edited by Juaz on 28 Jul 2007, 18:36, edited 2 times in total.
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Senior Manager
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Re: Tough P&C Question [#permalink]
28 Jul 2007, 15:33
Himalayan wrote: ajay_gmat wrote: A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3
(ii) 4
(iii) 5
(iv) 6
(v) 8 A. i would say 3. (jean + Paul) + (either joan or jessica) = 2 (paul) + (joan and jessica) (i.e. not stuart and jean) = 1 so total = 3 i didn't do this the formal way, but there are 4 groups
S A B
P J A
P J B
P A B
it states jean needs to be with paul, but not the other way around
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Director
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Re: Tough P&C Question [#permalink]
28 Jul 2007, 16:04
anonymousegmat wrote: Himalayan wrote: ajay_gmat wrote: A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3
(ii) 4
(iii) 5
(iv) 6
(v) 8 A. i would say 3. (jean + Paul) + (either joan or jessica) = 2 (paul) + (joan and jessica) (i.e. not stuart and jean) = 1 so total = 3 i didn't do this the formal way, but there are 4 groups S A B P J A P J B P A B it states jean needs to be with paul, but not the other way around you are right.
i also got 4 when i used formuala but thought should use this approach to minimize the risk. hmmmmmmm fell on the trap. toooo sad!!!!!!
yah agree with 4.
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Senior Manager
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Re: Tough P&C Question [#permalink]
28 Jul 2007, 21:03
Himalayan wrote: ajay_gmat wrote: A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3
(ii) 4
(iii) 5
(iv) 6
(v) 8 A. i would say 3. (jean + Paul) + (either joan or jessica) = 2 (paul) + (joan and jessica) (i.e. not stuart and jean) = 1 so total = 3 What about Stuart, Jessica and Joan ?
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Re: Tough P&C Question
[#permalink]
28 Jul 2007, 21:03
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