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GMAT Club Test Q [#permalink]
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21 May 2010, 13:52
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If X, Y,and Z are positive integers, how many different ordered sets(X,Y,Z) are possible such that X+Y+Z=5 ? A. 3 B. 4 C. 5 D. 6 E. 8 == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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Re: GMAT Club Test Q [#permalink]
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21 May 2010, 17:36
f X, Y,and Z are positive integers, how many different ordered sets(X,Y,Z) are possible such that X+Y+Z=5 ?
A. 3 B. 4 C. 5 D. 6 E. 8
Answer is D
(1,2,2) ( 2,2,1),(2,1,2) (1,1,3) (3,3,1),(1,3,1)



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Re: GMAT Club Test Q [#permalink]
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21 May 2010, 20:33
nitishmahajan wrote: If X, Y,and Z are positive integers, how many different ordered sets(X,Y,Z) are possible such that X+Y+Z=5 ?
A. 3 B. 4 C. 5 D. 6 E. 8 X,Y,Z >0 Lets start from X =1 1,1,3 1,2,2 1,3,1 2,1,2(repeat) 2,2,1(repeat) 3,1,1(repeat) We are talking about ordered sets and I haven't come across this on my GMAT preparation, so I am not sure if we should remove the repeated sets or not. So answer is either 3 or 6 i.e A or D I would go with D because ordered sets means that the order does matter.



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Re: GMAT Club Test Q [#permalink]
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07 Jul 2010, 19:49
As numbers are positive there are two possible outcomes:
1,2,2 and 1,1,3
Number of ways 1,2,2 can be arranged is: (1,2,2), (2,1,2) and (2,2,1) i.e. 3P3 So, 3P3 (1 + 1) = 3*2 = 6
Correct answer is D



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Re: GMAT Club Test Q [#permalink]
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16 Jul 2010, 02:45
two sets will be (1,2,2) and (1,1,3)
for each set arrangement possible will be 3! / 2! = 3
so total arrangements will be 3*2 = 6



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Re: GMAT Club Test Q [#permalink]
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22 Aug 2017, 21:52
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