November 18, 2018 November 18, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. November 18th, 7 AM PST November 20, 2018 November 20, 2018 09:00 AM PST 10:00 AM PST The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50621

Seven children — A, B, C, D, E, F, and G — are going to sit in seven c
[#permalink]
Show Tags
04 Mar 2015, 03:36
Question Stats:
71% (02:04) correct 30% (02:23) wrong based on 305 sessions
HideShow timer Statistics
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children? A. 600 B. 720 C. 1440 D. 4320 E. 4800 Kudos for a correct solution.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
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




Math Expert
Joined: 02 Sep 2009
Posts: 50621

Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c
[#permalink]
Show Tags
08 Mar 2015, 13:43
Bunuel wrote: Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?
A. 600 B. 720 C. 1440 D. 4320 E. 4800
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:This problem is a notch harder than anything you are likely to see on the GMAT. If you can master the principles in this problem, you certainly will be able to handle almost any problem the GMAT could concoct. First, consider the restriction of A & B. As with problem #1 above, there are 12 possibilities for A & B, counting both position and order. Now, put the other five in any order — that’s 120 possibilities, for a total number of configurations of 1440. That number does not take into account the restriction with C. Think about those 1440 configurations. In exactly half of them, C will be to the right of A & B, and exactly half, C will be to the left of A & B. Therefore, in (1/2)*1440 = 720 configurations, C will be to the right of A & B. 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



Senior Manager
Joined: 04 Aug 2010
Posts: 306
Schools: Dartmouth College

Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c
[#permalink]
Show Tags
06 Sep 2018, 05:05
Bunuel wrote: Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?
A. 600 B. 720 C. 1440 D. 4320 E. 4800 Since A and B must occupy adjacent positions, consider AB a single element in the arrangement. The number of ways to arrange the 6 elements AB, C, D, E, F and G = 6! = 720. In 1/2 of these arrangements, C will be to the LEFT of AB. In the remaining 1/2 of these arrangements,C will be to the RIGHT of AB. Thus, the number of arrangements in which C is to the right of AB = (1/2)(720). Since AB can switch to BA  doubling the total number of possible arrangements  we multiply by 2: (2)(1/2)(720) = 720.
_________________
GMAT and GRE Tutor Over 1800 followers Click here to learn more GMATGuruNY@gmail.com New York, NY If you find one of my posts helpful, please take a moment to click on the "Kudos" icon. Available for tutoring in NYC and longdistance. For more information, please email me at GMATGuruNY@gmail.com.




Math Expert
Joined: 02 Sep 2009
Posts: 50621

Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c
[#permalink]
Show Tags
04 Mar 2015, 03:38
Bunuel wrote: Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?
A. 600 B. 720 C. 1440 D. 4320 E. 4800
Kudos for a correct solution. Similar yet different questions: sixchildrenabcdeandfaregoingtositinsixchairsi194098.htmlsevenchildrenabcdefandgaregoingtositinsevenc194097.htmlsevenchildrenabcdefandgaregoingtositinsevenc194096.html
_________________
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



Manager
Joined: 20 Aug 2015
Posts: 96
Location: India
GPA: 3

Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c
[#permalink]
Show Tags
21 Oct 2015, 03:04
A, B, C, D, E, F, G seven children, of which A & B must sit next to each other. Considering them as one X (A, B), we have X, C, D, E, F, G. These can be arranged in 6! ways. But A,B can arrange themselves in 2! ways. So a total of 6!*2! ways = 1440. Since in exactly half of them, C will be to the right of A & B, and exactly half, C will be to the left of A & B, therefore 1440/2 = 720.
B is the right answer.



Director
Joined: 23 Jan 2013
Posts: 570

Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c
[#permalink]
Show Tags
22 Oct 2015, 03:11
it is
(6!*2)/2=6!=720
B



Senior Manager
Joined: 24 Oct 2016
Posts: 274
Location: India
Concentration: Finance, International Business
GPA: 3.96
WE: Human Resources (Retail Banking)

Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c
[#permalink]
Show Tags
08 Jul 2017, 04:58
the answer must be B say AB*5*4*3*2*1=120 4*AB*4*3*2*1=96 4*3*AB*3*2*1=72 4*3*2*AB*2*1=48 4*3*2*1*AB*1=24 total = 360 ways now as we also know that AB can be rearranged in every case in 2 ways so 360*2=720 ways
if you liked it please press kudos



Intern
Joined: 14 Jun 2017
Posts: 19

Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c
[#permalink]
Show Tags
21 Aug 2017, 22:00
Bunuel wrote: Bunuel wrote: Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?
A. 600 B. 720 C. 1440 D. 4320 E. 4800
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:This problem is a notch harder than anything you are likely to see on the GMAT. If you can master the principles in this problem, you certainly will be able to handle almost any problem the GMAT could concoct. First, consider the restriction of A & B. As with problem #1 above, there are 12 possibilities for A & B, counting both position and order. Now, put the other five in any order — that’s 120 possibilities, for a total number of configurations of 1440. That number does not take into account the restriction with C. Think about those 1440 configurations. In exactly half of them, C will be to the right of A & B, and exactly half, C will be to the left of A & B. Therefore, in (1/2)*1440 = 720 configurations, C will be to the right of A & B. Answer = B Hi, I am not quite sure i get how you come to the following conclusion "Think about those 1440 configurations. In exactly half of them, C will be to the right of A & B, and exactly half, C will be to the left of A & B"



Intern
Joined: 08 Jun 2015
Posts: 20
Location: United States
Concentration: Finance, Economics
WE: Engineering (Other)

Seven children — A, B, C, D, E, F, and G — are going to sit in seven c
[#permalink]
Show Tags
22 Aug 2017, 00:40
Hey Bunuel, can you please check this approach. Let 7 seats be 1 2 3 4 5 6 7 Condition : Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. So let this be the right side of A & B > 1 2 3 4 5 6 7. Now considering A & B as a single unit, we have 6 seats for them. (Omitting the number 1 seat as if A & B starts sitting from seat number 1, C can't be seated to their right). So we have 6C2 X 2! = 30. Since 3 out of 7 seats are filled as C, A, B needs to be seated in this particular order always. Out of remaining 4 seats 4 members can be arranged as 4! = 24. Hence answer would be 24 X 30 = 720.



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2700
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c
[#permalink]
Show Tags
06 Sep 2018, 03:43
Bunuel wrote: Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?
A. 600 B. 720 C. 1440 D. 4320 E. 4800
Kudos for a correct solution. With A and B together as one entity total arrangements of 6 entities = 6!*2! = 720*2 in half of the case C will sit to the right of A and B and in other half cases C will sit to the left of A and B Hence desired outcomes = (1/2)*6!*2! = 720 Answer: option B
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Intern
Joined: 21 Jan 2017
Posts: 33

Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c
[#permalink]
Show Tags
10 Sep 2018, 00:49
Bunuel wrote: Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?
A. 600 B. 720 C. 1440 D. 4320 E. 4800
Kudos for a correct solution. I solved it this way: Consider 7 seats. x x x x x x x x Place A and B together. case 1: ABxxxxx Here C can be arranged in 5 ways. The remaining children in 4! ways. A and B in 2! ways. So total: 2!.5.4! case 2: xABxxxx Here C can be arranged in 4 ways. The remaining children in 4! ways. A and B in 2! ways. So total: 2!.4.4! Similarly we can have : case 3: xxABxxx = 2!.3.4! case 4: xxxABxx = 2!.2.4! case 5:xxxxABx = 2!.1.4! adding all cases we get: 2!.4!(5+4+3+2+1) =2.24.5.3 =720 (B) (I followed this approach as the total combinations when AB taken as a unit and dividing by 2 didn't strike my brain :D , so found this alternate approach. Hit kudos if you like.) Thanks, Uma




Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c &nbs
[#permalink]
10 Sep 2018, 00:49






