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# A committee of 3 students has to be formed. The are five

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A committee of 3 students has to be formed. The are five [#permalink]

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28 Jul 2007, 12:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Senior Manager
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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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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.
Director
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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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28 Jul 2007, 14:54
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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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
Director
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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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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 ?
Re: Tough P&C Question   [#permalink] 28 Jul 2007, 21:03
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# A committee of 3 students has to be formed. The are five

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