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# GMAT Club Test Q

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Senior Manager
Joined: 25 Jun 2009
Posts: 302

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21 May 2010, 13:52
00:00

Difficulty:

(N/A)

Question Stats:

40% (01:59) correct 60% (00:39) wrong based on 11 sessions

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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
Intern
Joined: 21 Mar 2010
Posts: 17
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

(1,2,2) ( 2,2,1),(2,1,2)
(1,1,3) (3,3,1),(1,3,1)
Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 327
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.
Intern
Joined: 06 Jul 2010
Posts: 18
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

Manager
Joined: 12 Jun 2007
Posts: 125
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
Re: GMAT Club Test Q   [#permalink] 16 Jul 2010, 02:45
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