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# The principle of a school wants to arrange 5 students on the

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Senior Manager
Joined: 30 Aug 2003
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The principle of a school wants to arrange 5 students on the [#permalink]

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25 Jan 2004, 12:33
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The principle of a school wants to arrange 5 students on the platform such that the boy ROBERT occupies the second position and that NELLIE is always adjacents to the girl MELANIE. How many such arrangements are possible?

a.2
b.8
c.4
d.16
e.11
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

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Manager
Joined: 26 Dec 2003
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25 Jan 2004, 16:36
Sunnyboy u r back, I missed u. . I got 8 but I did the traditional way, R is always the 2nd position and N,M never occupy the 1st or 2nd position, so

ARNMB, ARMNB, BRNMA, BRMNA, ARBNM, ARBMN, BRANM, BRAMN

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Director
Joined: 03 Jul 2003
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26 Jan 2004, 11:49
By for the best combination question I've seen .
Luckly they haven't given 12 as one of the answer choice.
Otherwise, it would have been doomed.

Another classic example, why sometimes farmulas won't work.

My Anawer also 8.

Thanks a lot folks,

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Intern
Joined: 13 Jan 2004
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26 Jan 2004, 12:47
Draw it out

__ R __ __ __

How many options to place M and N ? only 2, and that is times 2! for their internal order.

Now, how many options to place A ? two places left, how many options to place B ??? only 1.

2x2!x2x1 = 2x2x2x1 = 8.

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Senior Manager
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26 Jan 2004, 14:58

_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

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Senior Manager
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29 Jan 2004, 10:39
sunniboy007 wrote:
:cool

The question needs to clarify the meaning of "second position" whther it is second from either end or just from the left side. The answer changes depending on this clarification. 8 is answer only if we assume that the second position is from LEFT. If assume the second position from either end, the answer would be 16.

I am not sure whether an ETS question will make this clarification or not.

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Manager
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29 Jan 2004, 11:18
I came up with 8.

the way I did was R position is fix so 4 positions remaining

out of 4 N&M can be arrenged in 6 ways with the given condition and the rest 2 can be in 2! with total of 8 ways.

please correct me if i am wrong in my approach. These Permutations, Combinations and probablility are my biggest headaches. Any suggestions on the study/practice materials of those

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29 Jan 2004, 11:18
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