Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 18 08:00 AM PDT  09:00 AM PDT Learn an intuitive, systematic approach that will maximize your success on Fillintheblank GMAT CR Questions. Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss!
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 06 Aug 2011
Posts: 318

In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
07 Jan 2014, 05:54
Question Stats:
42% (02:40) correct 58% (02:55) wrong based on 173 sessions
HideShow timer Statistics
In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing on line if each of Alice, Benjamin, Charlene must be on the line before each of Frederick, Gale, and Harold? A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !




Math Expert
Joined: 02 Sep 2009
Posts: 58453

In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
07 Jan 2014, 06:17
sanjoo wrote: In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing on line if each of Alice, Benjamin, Charlene must be on the line before each of Frederick, Gale, and Harold?
A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268 We want A, B, and C to be before F, G and H. Let's deal with this constraint first: consider A, B, and C as one unit {ABC} and F, G and H also as one unit {FGH}. According to the condition we need them to be arranged as {ABC}{FGH}. A, B, and C, within their unit can be arranged in 3! ways, similarly F, G and H , within their unit can also be arranged in 3! ways. So, we have a line with 6 people XXXXXX. D can take any place in the line so D has 7 options: *X*X*X*X*X*X*. Now, we have a line with 7 people and E can take any place there, so E has 8 options: *X*X*X*X*X*X*X*. Total = 3!*3!*7*8 = 2,016. Answer: D. Or: Place D and E first: 8 options for D, 7 options for E. We are left with 6 empty places. A, B and C must take first 3 places and F, G, H must take the remaining three. A, B and C can be arranged in their places in 3! ways. The same for F, G, and H. Total = 8*7*3!*3! = 2,016. Answer: D.
_________________




Intern
Joined: 17 Oct 2013
Posts: 40
GMAT Date: 02042014

Re: In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
07 Jan 2014, 06:54
sanjoo wrote: In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing on line if each of Alice, Benjamin, Charlene must be on the line before each of Frederick, Gale, and Harold?
A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268 first d,e can come at any point,=>8c2 * 2 next all three a,b,c should come before f,g,h so take all a,b,c as one unit and f,g,h as one unit. each of which can be written in 3! ways, hence total no of ways of arrangement are 8c2*2*3!*3!=2016



Senior Manager
Joined: 06 Aug 2011
Posts: 318

Re: In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
07 Jan 2014, 08:43
Thanks bharat and Bunuel.. Bt m not geting that 3! for abc and 3! fgh .. how can we say that abc will come before fgh?? Above both solution shows that abc will sit togather and fgh will sit togather.. d and will sit anywhere.. I think above solution is also for this question too.. Q: In how many different ways can eight people a,b,c,d,e,f,g and h, can be standing on line that abc and fgh will always stand togather? If i m wrong correct me
_________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !



Math Expert
Joined: 02 Sep 2009
Posts: 58453

In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
07 Jan 2014, 08:51
sanjoo wrote: Thanks bharat and Bunuel.. Bt m not geting that 3! for abc and 3! fgh .. how can we say that abc will come before fgh?? Above both solution shows that abc will sit togather and fgh will sit togather.. d and will sit anywhere.. I think above solution is also for this question too.. Q: In how many different ways can eight people a,b,c,d,e,f,g and h, can be standing on line that abc and fgh will always stand togather? If i m wrong correct me In the first approach {ABC} is before {FGH} because we consider only {ABC}{FGH} ordering and not {FGH}{ABC}. In the second approach A, B and C take first 3 places and F, G, H take the remaining 3. So, here A, B and C are also before F, G, and H. Hope it's clear.
_________________



Intern
Joined: 17 Oct 2013
Posts: 40
GMAT Date: 02042014

Re: In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
07 Jan 2014, 10:42
sanjoo wrote: Thanks bharat and Bunuel.. Bt m not geting that 3! for abc and 3! fgh .. how can we say that abc will come before fgh?? Above both solution shows that abc will sit togather and fgh will sit togather.. d and will sit anywhere.. I think above solution is also for this question too.. Q: In how many different ways can eight people a,b,c,d,e,f,g and h, can be standing on line that abc and fgh will always stand togather? If i m wrong correct me suppose number the seats as below 1 2 3 4 5 6 7 8 it is given that all three abc are before all three fgh, we first choose 2 seats randomly for d,e if we choose 1,4 for d,e.. 2,3,5 should be filled by abc which are of 3!ways and 6,7,8 should be filled with fgh in 3! ways hence a,b,c and f,g,h are not together but seems together because of the formula



Senior Manager
Joined: 06 Aug 2011
Posts: 318

Re: In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
07 Jan 2014, 12:07
Ok Bharat And Bunuel.. i got that.. Thanks alot
_________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !



Intern
Joined: 29 Dec 2012
Posts: 23
GMAT 1: 680 Q45 V38 GMAT 2: 690 Q47 V38
GPA: 3.69

Re: In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
22 Jul 2014, 17:57
What is wrong with my methodology below:
8 spots in line: _ _ _ _ _ _ _ _
ABC must be before DEF so possibilities are:
1) _ _ _ D E F _ _ For this option, ABC must occupy first three spots, and 3!*3!*2! = 72 ways 2) _ _ _ _ D E F _ For this option, ABC must occupy 3 of the first 4 slots, and D and E can be wherever, so 4!*2!*3! = 288 ways 3) _ _ _ _ _ D E F For this option, the remaining 5 people can be arranged in 5! ways = 720 ways
For a total of 1080 ways. Obviously this is wrong but I can't see why...



Math Expert
Joined: 02 Sep 2009
Posts: 58453

In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
23 Jul 2014, 02:43
vaj18psu wrote: What is wrong with my methodology below:
8 spots in line: _ _ _ _ _ _ _ _
ABC must be before DEF so possibilities are:
1) _ _ _ D E F _ _ For this option, ABC must occupy first three spots, and 3!*3!*2! = 72 ways 2) _ _ _ _ D E F _ For this option, ABC must occupy 3 of the first 4 slots, and D and E can be wherever, so 4!*2!*3! = 288 ways 3) _ _ _ _ _ D E F For this option, the remaining 5 people can be arranged in 5! ways = 720 ways
For a total of 1080 ways. Obviously this is wrong but I can't see why... The problem with your solution is that D, E and F are not necessary to be adjacent. G and/or H could be between them. This will increase the number of arrangements for each case.
_________________



Board of Directors
Joined: 17 Jul 2014
Posts: 2509
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
09 Jan 2016, 12:36
well, to be honest, I was lucky to pick the right answer.... i started this way: we have few options ABC D E FGH = 3!*5C1*4C1*3! = 6*5*4*6 = 720 D ABC E FGH = 5C1*3!*4C1*3! = 5*6*4*6 = 720 E ABC D FGH = 5C1*3!*4C1*3! = 5*6*4*6 = 720 D E ABC FGH = 5C1*4C1*3!*3! = 5*4*6*6 = 720. i get 2880. way to much. suppose we remove one of 720, we get 2160  still no match. where I did the mistake? ok, now I see. it might be ABC FGH D E or ABC D FGH E etc. or might be that ABC or FGH not one next to other... tricky question.



Intern
Joined: 13 Jun 2011
Posts: 21

Re: In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
26 Jan 2016, 01:02
Bunuel wrote: sanjoo wrote: In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing on line if each of Alice, Benjamin, Charlene must be on the line before each of Frederick, Gale, and Harold?
A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268 We want A, B, and C to be before F, G and H. Let's deal with this constraint first: consider A, B, and C as one unit {ABC} and F, G and H also as one unit {FGH}. According to the condition we need them to be arranged as {ABC}{FGH}. A, B, and C, within their unit can be arranged in 3! ways, similarly F, G and H , within their unit can also be arranged in 3! ways. So, we have a line with 6 people XXXXXX. D can take any place in the line so D has 7 options: *X*X*X*X*X*X*. Now, we have a line with 7 people and E can take any place there, so E has 8 options: *X*X*X*X*X*X*X*. Total = 3!*3!*7*8 = 2,016. Answer: D. Or: Place D and E first: 8 options for D, 7 options for E. We are left with 6 empty places. A, B and C must take first 3 places and F, G, H must take the remaining three. A, B and C can be arranged in their places in 3! ways. The same for F, G, and H. Total = 8*7*3!*3! = 2,016. Answer: D. Hi Brunel.. I am getting different ans.please tell me what is wrong with my approach 1)there are 8 spot.A,B,C,D,E AS ONE Subset and they are coming before F,G,H so can be arranged in 5! now considering F,G,H As one subset we can arrange them in 3! way..=so total is 5!*3! 2)another way 8 people can sit when ABC comes first and remaining 5 people can be in other subset so in total we can arrange them in 3!*5! so toal ways=A+B=720+720



Math Expert
Joined: 02 Aug 2009
Posts: 7978

Re: In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
18 May 2016, 20:35
sanjoo wrote: In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing on line if each of Alice, Benjamin, Charlene must be on the line before each of Frederick, Gale, and Harold?
A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268 In response to a PM.... To answer these Qs, we have to find how many of the ways would be invalid...there are 6 people  A,B,C,F,H and G ... they can be arranged in 6! ways... BUT among them ONLY valid solutions are when B,C,A, are ahead of F,G,H..so they can be arranged in such scenario in 3!*3! ways.... so the fraction of TOTAL that will be VALID is 3!*3!/6! = 20....so answer = 8!/20 = 2016 D
_________________



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15281
Location: United States (CA)

Re: In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
09 Mar 2018, 15:47
Hi All, This is a quirky permutation question. I think that the intent of the prompt is that we have a line of 8 people; from beginningtoend of that line, A, B and C must appear before we see F, G and H. This does NOT mean that A has to be first in line though. Here's how the math "works"… D and E are "floaters"  they can technically appear anywhere  so let's deal with them first: D can be in any of the 8 spots. Once we put D in a spot…. E can be in any of the 7 remaining spots… (8)(7) = 56 Now let's deal with the first group of 3: A, B and C. These three people can appear in any order, as long as they ALL appear before F, G and H. We can arrange A, B and C in 6 different ways: ABC ACB BAC BCA CAB CBA The same can be said for F, G and H: 6 different ways. (56)(6)(6) = 2016 If this were an official GMAT question, then the answers would likely be in ascending order. As you've listed them though… Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★



Director
Joined: 17 Dec 2012
Posts: 626
Location: India

Re: In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
28 Mar 2018, 22:16
sanjoo wrote: In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing on line if each of Alice, Benjamin, Charlene must be on the line before each of Frederick, Gale, and Harold?
A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268 Leftmost arrangement of the non constraint element is DEABCFGH Using formula, the number of permutations is 2!*3!*3!*(7+6+5+4+3+2+1)=72*28=2016 For explanation of the formula see link below
_________________
Srinivasan Vaidyaraman Sravna Test Prep http://www.sravnatestprep.comHolistic and Systematic Approach



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
WE: Education (Education)

In how many different orders can the people Alice, Benjamin,
[#permalink]
Show Tags
23 Jul 2019, 08:19
sanjoo wrote: In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing on line if each of Alice, Benjamin, Charlene must be on the line before each of Frederick, Gale, and Harold?
A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268 Let their positions are _ _ _ _ _ _ _ _ for 8 people since there is no condition applied on D and E so they can take any 2 of these 8 positions and arrange themselves in = 8C2*2! ways Now, 6 positions are left vacant which need to be occupied by ABCFGH but first three vacant positions must be occupied by A, B and C only in any order = 3! ways and next three vacant positions must be occupied by F, G and H only in any order = 3! ways So, Total Favourable arrangements = 8C2*2!*3!*3! = 2016 Answer: Option D To attend the !!!FREE!!! Live class on Permutation and combination on 24th July 2019 @9 PM IST by clicking and registering on the LINK
_________________
Prosper!!!GMATinsightBhoopendra 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 Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION




In how many different orders can the people Alice, Benjamin,
[#permalink]
23 Jul 2019, 08:19






