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

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

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New post 26 Jul 2014, 12:19
I just get the feeling that all these combinations are really misleading and unecessary to solve this one. Correct answer is 5 not 4 as official answer.

1 possible way - when Saul & Paul are out you have Jessica, Jane and Joan
3 ways when Sean is added to committe and combines: 1. Sean, Jessica, Joan 2. sean, jessica, jane & 3. Sean, joan, jane
1 way when paul is in the comitte - sean joan jessica

Adding those up it's 5! Am i crazy or what?

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

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New post 26 Jul 2014, 12:33
oanaandries wrote:
I just get the feeling that all these combinations are really misleading and unecessary to solve this one. Correct answer is 5 not 4 as official answer.

1 possible way - when Saul & Paul are out you have Jessica, Jane and Joan
3 ways when Sean is added to committe and combines: 1. Sean, Jessica, Joan 2. sean, jessica, jane & 3. Sean, joan, jane
1 way when paul is in the comitte - sean joan jessica

Adding those up it's 5! Am i crazy or what?

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The committees in red are not possible: Jane cannot be in the committee without Paul.

There are only 4 committees possible:
{Jane, Paul, Joan}
{Jane, Paul, Jessica}
{Paul, Jane, Jessica}
{Stuart, Joan, Jessica}

Please read the whole thread and try to understand the approaches presented there...
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Re: A committee of three students has to be formed. There are  [#permalink]

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New post 05 Dec 2014, 23:35
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
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Re: A committee of three students has to be formed. There are  [#permalink]

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New post 06 Dec 2014, 04:02
1
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.
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Re: A committee of three students has to be formed. There are  [#permalink]

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New post 06 Dec 2014, 05:40
Thanks Manpreet now I got the twist... It was confusing..
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Re: A committee of three students has to be formed. There are  [#permalink]

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New post 13 Mar 2016, 06:39
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?

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

m10 q25


Given: A, B, C, D, AND E
C and D are not together
If A, then C
The committee consists of three
with A we have to include C; the third person can be B or E => (ACB) OR (ACE)
with B , we have to choose from C, D, E; in this C and D are not together CE or DE=>(BCE) or (BDE)
with C, we have to choose from D and E => not possible
Therefore, there are only 4 possibilities.
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Re: A committee of three students has to be formed. There are  [#permalink]

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New post 16 Mar 2016, 19:13
There are two cases
1. When Jane serves in the committee. That implies Paul is in the committee. The remaining person can be chosen in 2 ways.

2. When Jane does not serve. In this case either Paul or Stuart can be the first member. There is just one way of selecting the remaining 2 members in each case. So a total of 2 ways when Jane does not serve

Total = 2+2=4
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Re: A committee of three students has to be formed. There are  [#permalink]

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New post 11 Jun 2018, 09:22
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?

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

m10 q25


Three Possible cases:

i)Committee has Paul & Jane, hence the third person can be selected is out of Joan & Jessica = 2 ways
ii)Committee has Paul but no Jane, hence the second person & third person can be selected out of again, Joan & Jessica, = 1 way
iii)Committee has Stuart, hence no Paul & hence no Jane, only Joan & Jessica can be selected = 1 way

Total # ways to form the committee = 2+1+1 = 4

Answer B.

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

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New post 11 Jun 2018, 09:44
Guys I think you make it more complicate

Here is my approach:
Without restrictions - we can make 5C3=10 committees

Subtract Paul's restrictions:
1) Paul, Stewart, Jane
2) Paul, Stewart, Joan
3) Paul, Stewart, Jessica

Subtract Jane's restrictions:
1) Jane, Joan, Stuart
2)Jane, Joan, Jessica
3) Jane, Stuart, Jessica

only 4 committees are left
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Re: A committee of three students has to be formed. There are  [#permalink]

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New post 13 Jun 2018, 06:09
1
It seemed to me very easy question (maybe because not a big number of options are there).
Since Paul and Stuart cannot be together and Jane will not be in committee without Paul, lets consider first Paul.
1.Paul + Jane + Joan
2.Joan+Paul+jess
3.paul+jess+jane
Now, consider stuard,
4.Stuart+ jess+jane
4 in total, answer is B
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Re: A committee of three students has to be formed. There are  [#permalink]

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Re: A committee of three students has to be formed. There are   [#permalink] 11 Jul 2019, 17:14

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