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Re M1025
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15 Sep 2014, 23:42
Official Solution: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 # of committees with Paul is \(C^2_3=3\), since we choose remaining 2 members out of 3 people: Jane, Joan and Jessica (so everyone but Stuart); # of committees with Stuart is \(C^2_2=1\), since we choose remaining 2 members out of 2 people: Joan and Jessica (so everyone but Paul and Jane); Total: \(3+1=4\). Notice that no committees are possible if neither Paul nor Stuart are there, since no Paul means no Jane too, so only 2 members are left (Joan and Jessica) which cannot form 3member committee. Answer: B
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Re: M1025
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31 Mar 2016, 09:50
Correct me if I'm mistaken; in the solution it should be 1 committee with Paul and 3 with Stuart. thanks



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Re: M1025
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31 Mar 2016, 11:17
Quote: # of committees with Paul is C 2 3 =3 C32=3 , since we choose remaining 2 members out of 3 people: Jane, Joan and Jessica (so everyone but Stuart);
# of committees with Stuart is C 2 2 =1 C22=1 , since we choose remaining 2 members out of 2 people: Joan and Jessica (so everyone but Paul and Jane);
Isn't it reversed? Paul 1 and Stuart 3? Thank you



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Re: M1025
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31 Mar 2016, 11:27
Avigano wrote: Quote: 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?
# of committees with Paul is C 2 3 =3 C32=3 , since we choose remaining 2 members out of 3 people: Jane, Joan and Jessica (so everyone but Stuart);
# of committees with Stuart is C 2 2 =1 C22=1 , since we choose remaining 2 members out of 2 people: Joan and Jessica (so everyone but Paul and Jane);
Isn't it reversed? Paul 1 and Stuart 3? Thank you It's not that hard to list the committees: {Paul, Jane, Joan} {Paul, Jane, Jessica} {Paul, Joan, Jessica} {Stuart, Joan and Jessica}
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Re: M1025
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31 Mar 2016, 11:36
I just realized that I read "Jane refuses to be in the committee without Paul".. as "Jane refuses to be in the committee WITH Paul" So yeah, well Sh"#$ happens! My ugly mistake, pardon me.



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Re: M1025
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20 May 2016, 05:12
Bunuel wrote: Official Solution:
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
# of committees with Paul is \(C^2_3=3\), since we choose remaining 2 members out of 3 people: Jane, Joan and Jessica (so everyone but Stuart); # of committees with Stuart is \(C^2_2=1\), since we choose remaining 2 members out of 2 people: Joan and Jessica (so everyone but Paul and Jane); Total: \(3+1=4\). Notice that no committees are possible if neither Paul nor Stuart are there, since no Paul means no Jane too, so only 2 members are left (Joan and Jessica) which cannot form 3member committee. Answer: B Hi Bunnel, I have worked as below : With paul we have total 4 members to form 3 students as committe i.e 4C3 = 4C1 = 4 with stuart we have only combination , 3C3 =1 total = 4+1 = 5 could you please help me where am i going wrong. Thanks.



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Re: M1025
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20 May 2016, 05:43
babuvgmat wrote: Bunuel wrote: Official Solution:
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
# of committees with Paul is \(C^2_3=3\), since we choose remaining 2 members out of 3 people: Jane, Joan and Jessica (so everyone but Stuart); # of committees with Stuart is \(C^2_2=1\), since we choose remaining 2 members out of 2 people: Joan and Jessica (so everyone but Paul and Jane); Total: \(3+1=4\). Notice that no committees are possible if neither Paul nor Stuart are there, since no Paul means no Jane too, so only 2 members are left (Joan and Jessica) which cannot form 3member committee. Answer: B Hi Bunnel, I have worked as below : With paul we have total 4 members to form 3 students as committe i.e 4C3 = 4C1 = 4with stuart we have only combination , 3C3 =1 total = 4+1 = 5 could you please help me where am i going wrong. Thanks. You are going wrong in the highlighted portion... If paul is there in the team, stuart cannot be there and we have to choose 2 out of remaining three = 3C2 = 3.. If paul is not there, jane will also be not there, there is ONLY one team possible 1C1=1..
total 3+1=4..whereas in the highlighted portion if you take 4C3, you are adding one scenario where paul is not there and Jane is there
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Re: M1025
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20 May 2016, 06:33
whereas in the highlighted portion if you take 4C3, you are adding one scenario where paul is not there and Jane is there[/quote] Ohh yeah .. this point i was missing . Thank you Bunnel



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Re: M1025
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20 May 2016, 06:46
babuvgmat wrote: whereas in the highlighted portion if you take 4C3, you are adding one scenario where paul is not there and Jane is there Ohh yeah .. this point i was missing . Thank you Bunnel [/quote] hi, Bunuel, you have so many great posts floating around that any post clearing any query is believed to be yours .....
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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?
I have worked in a different way :
1/ The total number of committees : 3 from 5 = \(\frac{(5*4*3)}{(3*2*1)}\)= 10
2/ The number of committees which break the rules : i.e. : the number of committees with Paul and Stuart, and the number of committees with Jane and not Paul
a) with Paul and Stuart : 1*1*3 = 3 committees
b) with Jane and not Paul : \(\frac{(1*3*2)}{(2*1)}\) = 3 committees as well
Finally, 10  3 3 = 4



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Re: M1025
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30 May 2016, 18:18
Alex75PAris 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?
I have worked in a different way :
1/ The total number of committees : 3 from 5 = \(\frac{(5*4*3)}{(3*2*1)}\)= 10
2/ The number of committees which break the rules : i.e. : the number of committees with Paul and Stuart, and the number of committees with Jane and not Paul
a) with Paul and Stuart : 1*1*3 = 3 committees
b) with Jane and not Paul : \(\frac{(1*3*2)}{(2*1)}\) = 3 committees as well
Finally, 10  3 3 = 4 Hi, Why did you divided 2*1 into 1*3*2 when you were computing numbers around Jane without Paul?



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Re: M1025
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06 Jul 2016, 04:43
Another approach:
Committees with Jane >> Paul is member, Stuart excluded >> Committee Possibilities = Possibilities for 3rd member: 2 (either Joan or Jessica)
Committees without Jane >> Paul cannot be with Stuart (enemies restriction) >> treat Paul + Stuart as 1 unit (ways to pick within unit = 2), need to pick 3 from 3 choices >> Committee Possibilities = (3!/3!) * 2 = 2
Total committee possibilities = 2 + 2 = 4



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Re: M1025
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02 Sep 2016, 09:54
Jane, Joan, Paul, Stuart, and Jessica jane and paul together ( if paul is there stuart is not there) so out of remaining three we can choose only 02 (jane, paul, jessica) or (jane,paul, joan)
if paul is not there jane is not there so 01 combination  (joan, stuart, jessica)
if paul is there stuart can not be there, also jane dosnt want to be without paul, she can be outside of the group too ( paul, jesscia, joan
total =4



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Re M1025
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05 Apr 2018, 09:21
I think this is a poorquality question and I don't agree with the explanation. when paul is a part of a committee then jane has to be part of that committee.so no of possible cases should be : paul,jane,jessica/joan. again when stuart is a part of that committee jane & paul wont be a part of that.so no of possible case is 1. so the total no of possible cases are :3.



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Re: M1025
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05 Apr 2018, 12:43
nilotpal1990 wrote: I think this is a poorquality question and I don't agree with the explanation. when paul is a part of a committee then jane has to be part of that committee.so no of possible cases should be : paul,jane,jessica/joan. again when stuart is a part of that committee jane & paul wont be a part of that.so no of possible case is 1. so the total no of possible cases are :3. It's not that hard to list the committees: {Paul, Jane, Joan} {Paul, Jane, Jessica} {Paul, Joan, Jessica} {Stuart, Joan and Jessica}
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Re: M1025
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04 Aug 2018, 01:10
A really good question! Took awhile to solve but nevertheless got it finally.
Total ways = \(5C3\) = 10 P(Paul + Stuart) = \(3C1 * 2C2 = 3\) P(Jane without Paul) = \(3C2 * \frac{2C1}{2} = 3\)
Answer is 1033 = 6



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Re: M1025
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26 Aug 2018, 04:30
hi bunuel, I solved it this way: if jane is on the committee then paul will be on committee so the last place can be filled in 2 ways. if stuart is on committee then we have to fill the remaining two places with joan and jessica in 2*1 ways.So total ways are 2+2=4 ways. Is this method correct?










