Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50007

Question Stats:
41% (01:48) correct 59% (01:45) wrong based on 117 sessions
HideShow timer Statistics



Math Expert
Joined: 02 Sep 2009
Posts: 50007

Re M1025
[#permalink]
Show Tags
16 Sep 2014, 00:42
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
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Current Student
Joined: 12 Nov 2015
Posts: 59
Location: Uruguay
Concentration: General Management
GMAT 1: 610 Q41 V32 GMAT 2: 620 Q45 V31 GMAT 3: 640 Q46 V32
GPA: 3.97

Re: M1025
[#permalink]
Show Tags
31 Mar 2016, 10:50
Correct me if I'm mistaken; in the solution it should be 1 committee with Paul and 3 with Stuart. thanks



Math Expert
Joined: 02 Sep 2009
Posts: 50007

Re: M1025
[#permalink]
Show Tags
31 Mar 2016, 11:34



Current Student
Joined: 12 Nov 2015
Posts: 59
Location: Uruguay
Concentration: General Management
GMAT 1: 610 Q41 V32 GMAT 2: 620 Q45 V31 GMAT 3: 640 Q46 V32
GPA: 3.97

Re: M1025
[#permalink]
Show Tags
31 Mar 2016, 12:17
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



Math Expert
Joined: 02 Sep 2009
Posts: 50007

Re: M1025
[#permalink]
Show Tags
31 Mar 2016, 12:27
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}
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Current Student
Joined: 12 Nov 2015
Posts: 59
Location: Uruguay
Concentration: General Management
GMAT 1: 610 Q41 V32 GMAT 2: 620 Q45 V31 GMAT 3: 640 Q46 V32
GPA: 3.97

Re: M1025
[#permalink]
Show Tags
31 Mar 2016, 12:36
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.



Intern
Joined: 10 Dec 2014
Posts: 39
Location: India
Concentration: Entrepreneurship, Operations
GPA: 4
WE: Operations (Consulting)

Re: M1025
[#permalink]
Show Tags
20 May 2016, 06:12
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.



Math Expert
Joined: 02 Aug 2009
Posts: 6967

Re: M1025
[#permalink]
Show Tags
20 May 2016, 06:43
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
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Intern
Joined: 10 Dec 2014
Posts: 39
Location: India
Concentration: Entrepreneurship, Operations
GPA: 4
WE: Operations (Consulting)

Re: M1025
[#permalink]
Show Tags
20 May 2016, 07:33
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



Math Expert
Joined: 02 Aug 2009
Posts: 6967

Re: M1025
[#permalink]
Show Tags
20 May 2016, 07:46
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 .....
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Manager
Joined: 16 Mar 2016
Posts: 129
Location: France
GPA: 3.25

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



Intern
Joined: 30 Jan 2015
Posts: 3

Re: M1025
[#permalink]
Show Tags
30 May 2016, 19:18
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?



Intern
Joined: 16 Feb 2015
Posts: 3

Re: M1025
[#permalink]
Show Tags
06 Jul 2016, 05:43
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



Manager
Joined: 17 Aug 2015
Posts: 110

Re: M1025
[#permalink]
Show Tags
02 Sep 2016, 10:54
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



Intern
Joined: 26 Jan 2018
Posts: 1

Re M1025
[#permalink]
Show Tags
05 Apr 2018, 10:21
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.



Math Expert
Joined: 02 Sep 2009
Posts: 50007

Re: M1025
[#permalink]
Show Tags
05 Apr 2018, 13:43
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}
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 20 Dec 2017
Posts: 35
Location: Singapore
GPA: 3.56

Re: M1025
[#permalink]
Show Tags
04 Aug 2018, 02:10
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



Intern
Joined: 09 May 2018
Posts: 7

Re: M1025
[#permalink]
Show Tags
26 Aug 2018, 05:30
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?










