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Senior Manager
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GMAT Club Test Q [#permalink] New post 21 May 2010, 12:52
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

40% (01:59) correct 60% (00:39) wrong based on 10 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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Re: GMAT Club Test Q [#permalink] New post 21 May 2010, 16:36
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)
Senior Manager
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Re: GMAT Club Test Q [#permalink] New post 21 May 2010, 19:33
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] New post 07 Jul 2010, 18:49
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] New post 16 Jul 2010, 01:45
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
Re: GMAT Club Test Q   [#permalink] 16 Jul 2010, 01:45
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