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There are 5 cars to be displayed in 5 parking spaces with

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Joined: 29 Aug 2005
Posts: 855

Kudos [?]: 487 [0], given: 7

There are 5 cars to be displayed in 5 parking spaces with [#permalink]

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New post 09 Nov 2008, 08:17
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction.
Of the 5 cars, 3 are red, 1 is blue, and 1 is yellow. If the cars are identical except for color, how
many different display arrangements of the 5 cars are possible?
A. 20
B. 25
C. 40
D. 60
E. 125

Kudos [?]: 487 [0], given: 7

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Joined: 30 Apr 2008
Posts: 1868

Kudos [?]: 615 [1], given: 32

Location: Oklahoma City
Schools: Hard Knocks
Re: GMAT Set 30 - 31 [#permalink]

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New post 09 Nov 2008, 08:27
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This is a permutations/combinations question.

First, disregarding the color, there can be 5! (5! = 120) arrangements. But we have to take into account the many ways the 3 red cards can be arranged among themselves that will not alter the actual color sequence of the arrangement. This is 3! (3! = 6) So that is 120 / 6 = 20

Answer A.

botirvoy wrote:
There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction.
Of the 5 cars, 3 are red, 1 is blue, and 1 is yellow. If the cars are identical except for color, how
many different display arrangements of the 5 cars are possible?
A. 20
B. 25
C. 40
D. 60
E. 125

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Kudos [?]: 615 [1], given: 32

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Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 842 [0], given: 19

Re: GMAT Set 30 - 31 [#permalink]

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New post 09 Nov 2008, 09:29
jallenmorris wrote:
This is a permutations/combinations question.

First, disregarding the color, there can be 5! (5! = 120) arrangements. But we have to take into account the many ways the 3 red cards can be arranged among themselves that will not alter the actual color sequence of the arrangement. This is 3! (3! = 6) So that is 120 / 6 = 20

Answer A.

botirvoy wrote:
There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction.
Of the 5 cars, 3 are red, 1 is blue, and 1 is yellow. If the cars are identical except for color, how
many different display arrangements of the 5 cars are possible?
A. 20
B. 25
C. 40
D. 60
E. 125



Nice one.

= 5!/3!
= 20
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Kudos [?]: 842 [0], given: 19

Re: GMAT Set 30 - 31   [#permalink] 09 Nov 2008, 09:29
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