GMAT Prep - Tough question: 3 of 7

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GMAT Prep - Tough question: 3 of 7

by GMATDUD » Mon Feb 04, 2008 3:41 pm
Hi Folks ,

another tough one : Any pointers would be greatly appreciated.

Thanks,
T
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solution

by tutonaranjo » Mon Feb 04, 2008 5:17 pm
this is a permutation problem

But the rule for arranging people in a round table is that it is n-1. This is because order matters and if not you would be counting the one specific permutation twice.

n=5, n-1=4
solution is= 1x2x3x4 = 24
Answer is 24

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by cris » Tue Feb 05, 2008 1:55 am
I dont understand what "two seating arrangements are considered different if the positions of the peole are different relative to each other".

This means that:

ABCDE will not be different from BACDE bc D is between C and E in both arrengements??

Wouldnt then the corret answer be A) 5?

:?:

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by Stuart@KaplanGMAT » Tue Feb 05, 2008 12:16 pm
cris wrote:I dont understand what "two seating arrangements are considered different if the positions of the peole are different relative to each other".

This means that:

ABCDE will not be different from BACDE bc D is between C and E in both arrengements??

Wouldnt then the corret answer be A) 5?

:?:
That extra line is in there because it's a circle arrangement problem. Basically, it's saying that it doesn't matter who sits in which seat, just where the people are relative to each other.

Let's call the seats 1, 2, 3, 4 and 5.

If we have A,B,C,D,E in seats 1-5, respectively, that would give the same relational arrangement as B,C,D,E,A in seats 1-5, respectively. Without the extra sentence in the question, we would count these as two separate arrangements.

(Also, the elimination of these duplicates is why the answer is 4! instead of 5!, which would be the solution for linearly arranging 5 people.)
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