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

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

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New post 16 Dec 2009, 22:53
2
4
00:00
A
B
C
D
E

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Question Stats:

84% (00:49) correct 16% (00:46) wrong based on 262 sessions

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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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Re: ARRANGEMENTS  [#permalink]

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New post 16 Dec 2009, 23:45
1
I'm going A.

I think you just think of it as 2 different cars and 5 empty spots, since the rest have to be red anyway. Then there's 5 spaces for the yellow car and four spaces for the blue one for each yellow car space. 4x5 = 20

Probably if I could remember the maths from permutations and combinations you'd end up with something like 5!/3!
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Re: ARRANGEMENTS  [#permalink]

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New post 17 Dec 2009, 02:19
1
2
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

I think the above logic is correct except one thing:

# of arrangements is \(\frac{5!}{3!}=20\).

But a car can be arranged facing TWO directions (within one particular parking space), so we should multiply 20 by 2, \(20*2=40\)

Answer: C.

Though I think that the question is quite ambiguous. We are told that "all the cars are facing the SAME direction", not that they all are facing ONE specific direction. That's why I think we should multiply by 2.

They can be arranged as
-->
-->
-->
-->
-->
20 combinations.

AND
<--
<--
<--
<--
<--
20 combinations.

Stem just says that they can can not be arranged for example like this:
<--
-->
-->
<--
<--
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Re: ARRANGEMENTS  [#permalink]

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New post 30 Dec 2013, 08:41
Bunuel 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

I think the above logic is correct except one thing:

# of arrangements is \(\frac{5!}{3!}=20\).

But a car can be arranged facing TWO directions (within one particular parking space), so we should multiply 20 by 2, \(20*2=40\)

Answer: C.

Though I think that the question is quite ambiguous. We are told that "all the cars are facing the SAME direction", not that they all are facing ONE specific direction. That's why I think we should multiply by 2.

They can be arranged as
-->
-->
-->
-->
-->
20 combinations.

AND
<--
<--
<--
<--
<--
20 combinations.

Stem just says that they can can not be arranged for example like this:
<--
-->
-->
<--
<--


Yup, Bunuel got it

I also forgot about multiplying by 2 but your point is valid

Cheers!
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Re: There are 5 cars to be displayed in 5 parking spaces with  [#permalink]

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New post 03 Aug 2018, 01:20
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: There are 5 cars to be displayed in 5 parking spaces with &nbs [#permalink] 03 Aug 2018, 01:20
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