GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Oct 2019, 20:00

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# In how many different orders can the people Alice, Benjamin,

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
2
33
00:00

Difficulty:

95% (hard)

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

_________________
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
7
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.

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.

_________________
##### General Discussion
Intern
Joined: 17 Oct 2013
Posts: 40
Schools: HEC Dec"18
GMAT Date: 02-04-2014
Re: In how many different orders can the people Alice, Benjamin,  [#permalink]

### Show Tags

07 Jan 2014, 06:54
2
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
2
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
2
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
Schools: HEC Dec"18
GMAT Date: 02-04-2014
Re: In how many different orders can the people Alice, Benjamin,  [#permalink]

### Show Tags

07 Jan 2014, 10:42
1
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
Schools: Tuck '17 (M)
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
GMAT 1: 650 Q49 V30
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.

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.

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
3
1
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/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15281
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
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 beginning-to-end 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…

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com

The 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.com

Holistic and Systematic Approach
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
Schools: Darden '21
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

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!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One 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
In how many different orders can the people Alice, Benjamin,   [#permalink] 23 Jul 2019, 08:19
Display posts from previous: Sort by