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

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Intern
Joined: 20 Sep 2008
Posts: 46
There are 5 cars to be displayed in 5 parking spaces with [#permalink]

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20 Sep 2008, 05:26
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
VP
Joined: 30 Jun 2008
Posts: 1026
Re: Cars Permutation [#permalink]

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20 Sep 2008, 05:58
is it A ?

5!/(3!*1!*1!) = 20
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"You have to find it. No one else can find it for you." - Bjorn Borg

SVP
Joined: 21 Jul 2006
Posts: 1508
Re: Cars Permutation [#permalink]

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20 Sep 2008, 06:01
jugolo1 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

Very weird question. I did 5! to get 120, but I can't find that in the answer choice. Is the question even complete? because I don't see any restrictions in the question. May be I'm missing something, but I'll be very interested to do know the answer to this question.
VP
Joined: 30 Jun 2008
Posts: 1026
Re: Cars Permutation [#permalink]

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20 Sep 2008, 06:02
tarek99 wrote:
jugolo1 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

Very weird question. I did 5! to get 120, but I can't find that in the answer choice. Is the question even complete? because I don't see any restrictions in the question. May be I'm missing something, but I'll be very interested to do know the answer to this question.

Isn't the COLOR restriction ??
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

SVP
Joined: 21 Jul 2006
Posts: 1508
Re: Cars Permutation [#permalink]

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20 Sep 2008, 06:03
amitdgr wrote:
is it A ?

5!/(3!*1!*1!) = 20

ahh, yes I agree with your approach. The only thing is that you don't have to divide the whole thing by also 1! and 1! because you're only required to divide by the repeating objects, which, in this case, are the 3 red cars.

I agree that the answer is A
VP
Joined: 30 Jun 2008
Posts: 1026
Re: Cars Permutation [#permalink]

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20 Sep 2008, 06:05
I think the question is similar to

what is the number of words that can be formed from the word RRRBY

Total no. of letters = 5, so total ways of arranging them = 5!

But R is repeated 3 times, B one time and Y one time

so it is 5!/(3!*1!*1!) = 20
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

SVP
Joined: 21 Jul 2006
Posts: 1508
Re: Cars Permutation [#permalink]

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20 Sep 2008, 06:13
amitdgr wrote:
I think the question is similar to

what is the number of words that can be formed from the word RRRBY

Total no. of letters = 5, so total ways of arranging them = 5!

But R is repeated 3 times, B one time and Y one time

so it is 5!/(3!*1!*1!) = 20

well true, but if you have only 1 object, such as B and Y in your example, they're not really repeating. You will get the same answer when you only divide by objects that are repeating...what you said isn't wrong, but your equation can be a lot simpler by just sticking to objects that are only repeating. B and Y aren't really repeating because they are unique objects. In your example: you can just do 5!/3! and still have the same answer.
VP
Joined: 30 Jun 2008
Posts: 1026
Re: Cars Permutation [#permalink]

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20 Sep 2008, 07:23
tarek99 wrote:
amitdgr wrote:
I think the question is similar to

what is the number of words that can be formed from the word RRRBY

Total no. of letters = 5, so total ways of arranging them = 5!

But R is repeated 3 times, B one time and Y one time

so it is 5!/(3!*1!*1!) = 20

well true, but if you have only 1 object, such as B and Y in your example, they're not really repeating. You will get the same answer when you only divide by objects that are repeating...what you said isn't wrong, but your equation can be a lot simpler by just sticking to objects that are only repeating. B and Y aren't really repeating because they are unique objects. In your example: you can just do 5!/3! and still have the same answer.

by explicitly showing 1! , I was just trying to be clear
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Re: Cars Permutation   [#permalink] 20 Sep 2008, 07:23
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# There are 5 cars to be displayed in 5 parking spaces with

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