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

Question

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

A. 1,008
B. 1,296
C. 1,512
D. 2,016
E. 2,268

Bunuel · Accepted Answer

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.

bharatdivvela · Answer

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

sanjoo · Answer

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

Bunuel · Answer

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.

bharatdivvela · Answer

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

sanjoo · Answer

Ok Bharat And Bunuel.. i got that.. Thanks alot

vaj18psu · Answer

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...

Bunuel · Answer

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.

mvictor · Answer

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.

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.

vaidhaichaturvedi · Answer

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

chetan2u · Answer

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

EMPOWERgmatRichC · Answer

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…

Final Answer:  D

GMAT assassins aren't born, they're made, 
Rich

SVaidyaraman · Answer

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

GMATinsight · Answer

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

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Rakhi08 · Answer

Hi  Bunuel

In this question why are we not using this logic --> i.e. out of total arrangements (8!) half the time ABC will come before FGH? so 8! / 2 ?
is it because we have D & E remaining?

Please help!

thanks in advance

EMPOWERgmatRichC · Answer

Hi Rakhi08,

Based on what you are asking, I think that you are thinking about a pattern that you may have seen from a different question and trying to apply it here.

IF you are considering just 2 letters (for example, A and F), and you arrange a series of 8 letters in a row, then HALF of the time, the A will be to the 'left' of the F. However, in this question, you need 3 letters (A, B and C) to ALL be to the 'left' of 3 other letters (F, G and H) and that does NOT happen half of the time.

Consider an example that does not include the two 'extra' letters (the D and the E). If we had ABCFGH, then the first 3 letters would have to be ABC (in some order) and the last 3 letters would have to be FGH (in some order). That would be (3!)(3!) = 36... However 6! = 720, so clearly far LESS than half of those 720 options fit what we would be looking for. Including those 2 additional letters increases the number of acceptable arrangements, but it does NOT increase the 'ratio' of acceptable arrangements to total arrangements (it's still just 1/20 of the total). 
 
GMAT assassins aren't born, they're made,
Rich

Oppenheimer1945 · Answer

_A_B_C_D_F_G_H_

7 spaces, ABC, FGH can be arranged in 3! ,3! ways

case 1) DE together= 7C1 2! 3! 3!

case 2) DE separate= 7C2 2! 3! 3#

Ans= 1512+504=2016

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

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