Permutation/Arrangement

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Permutation/Arrangement

by raj44 » Tue Aug 19, 2014 9:09 pm
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.1008
b.1296
c.2016
d.1512
e.2268

OA C
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by Brent@GMATPrepNow » Tue Aug 19, 2014 9:16 pm
raj44 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.1008
b.1296
c.2016
d.1512
e.2268

OA C
This question is poorly worded.
Does this mean that Alice must be ahead of Frederick, Gale, AND Harold, or just ahead of Frederick?
Likewise, does Benjamin need to be ahead of Frederick, Gale, AND Harold, or just ahead of Gale?

Cheers,
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by [email protected] » Wed Aug 20, 2014 12:19 am
Hi raj44,

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: C

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by GMATGuruNY » Wed Aug 20, 2014 2:26 am
raj44 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.1008
b.1296
c.2016
d.1512
e.2268
Number of options for D = 8. (Any of the 8 positions.)
Number of options for E = 7. (Any of the 7 remaining positions.)

A, B and C must stand leftward of F, G and H.
Implication:
Of the 6 positions left, A, B and C must occupy the 3 most leftward.
Number of options for A = 3. (Any of the 3 most leftward positions.)
Number of options for B = 2. (Either of the 2 remaining most leftward positions.)
Number of options for C = 1. (Only 1 of the most leftward positions remains.)

Only 3 positions remain for F, G and H.
Number of options for F = 3. (Any of the 3 remaining positions.)
Number of options for G = 2. (Either of the 2 remaining positions.)
Number of options for H = 1. (Only 1 position remains.)

To combine the options above, we multiply:
8*7*3*2*1*3*2*1 = 2016.

The correct answer is C.
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by raj44 » Wed Aug 20, 2014 4:37 am
Hi Brent,

Indeed the question wording is ambiguous! I interpreted that ABC must be arranged in such a way that they are always left/before to FGH.

Rajeev
Brent@GMATPrepNow wrote:
raj44 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.1008
b.1296
c.2016
d.1512
e.2268

OA C
This question is poorly worded.
Does this mean that Alice must be ahead of Frederick, Gale, AND Harold, or just ahead of Frederick?
Likewise, does Benjamin need to be ahead of Frederick, Gale, AND Harold, or just ahead of Gale?

Cheers,
Brent

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by raj44 » Wed Aug 20, 2014 4:40 am
Hi Rich,

You're right in figuring out the option order! :D I had deliberately interchanged the options - is it always true that options in gmat will be in ascending order?

Rajeev
[email protected] wrote:Hi raj44,

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: C

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by GMATinsight » Wed Aug 20, 2014 9:45 am
raj44 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.1008
b.1296
c.2016
d.1512
e.2268

OA C
The question only mentions the Constraint about positioning of ABC and FGH and it's silent on D and E who can sit anywhere.

Let's find out the places for D and E out of 8 Places = 8C2 = 28 ways
D and E can exchange their places with Each other in 2! ways

Therefore to Make D and E take 2 places, the total ways = 28 x 2 = 56 Ways

Let's say the seating arrangement now is
_ _ _ D _ E _ _

Now Since ABC have to be arranged before FGH therefore From Left first three places must be occupied by ABC which can happen in 3! ways = 6 ways

Similarly FGH can take last three places in 3! ways = 6 ways

Total ways to arrange them = 56 x 6 x 6 = 2016

Answer: Option C
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by [email protected] » Wed Aug 20, 2014 11:28 am
Hi Rajeev,

In the Quant section, whenever the answers are numbers, they're almost always in ascending order (answer A is smallest, answer E is largest). There are sometimes exceptions though (answers with exponents, but different bases, for example). This pattern can be beneficial, especially if you can determine that certain answers are "too big" or "too small", which allows you to eliminate options in the event that you have to guess.

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