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Senior Manager
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Question Stats:
33% (01:55) correct
66% (00:39) wrong based on 0 sessions
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
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Intern
Joined: 21 Mar 2010
Posts: 17
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Re: GMAT Club Test Q [#permalink]
 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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Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
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Re: GMAT Club Test Q [#permalink]
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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Intern
Joined: 06 Jul 2010
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Re: GMAT Club Test Q [#permalink]
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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Manager
Joined: 12 Jun 2007
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Re: GMAT Club Test Q [#permalink]
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]
16 Jul 2010, 02:45
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