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

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Manager
Joined: 02 Sep 2008
Posts: 100
There are 5 cars to be displayed in 5 parking spaces with  [#permalink]

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17 Mar 2009, 20:21
1
5
00:00

Difficulty:

(N/A)

Question Stats:

74% (00:43) correct 26% (00:31) wrong based on 233 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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Director
Joined: 25 Oct 2006
Posts: 551

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17 Mar 2009, 20:35
1
alpha_plus_gamma wrote:
5!/3! = 120/6 = 20
A

why you are not considering that cars could be aligned on different side too? Question says that all cars must be at same direction.
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Intern
Joined: 21 Jul 2006
Posts: 17

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24 Mar 2009, 15:29
Can somebody please explaine me how you got these numbers?
The way I did, obviously wrong was 5C3*5C1*5C1 = 250:(
What am I doing wrong? This perm/comb is kicking my butt!
Intern
Joined: 30 Jul 2009
Posts: 18

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08 Sep 2009, 13:15
1
1
matrix777 wrote:
Can somebody please explaine me how you got these numbers?
The way I did, obviously wrong was 5C3*5C1*5C1 = 250:(
What am I doing wrong? This perm/comb is kicking my butt!

I got 5!/3!, which equals 20. Since the cars are all facing the same direction, direction is irrelevant. 5P5 or 5! indicates the number of ways the 5 cars can be arranged. However, since three of the cars are red, we need to divide by 3! since an arrangement such as : R1, R2, R3, B, Y is the same as R2,R3,R1,B,Y or R3,R2,R1,B,Y.

Hopefully this makes sense.
Director
Joined: 01 Jan 2008
Posts: 593

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08 Sep 2009, 13:50
1
1
milind1979 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

number of ways to arrange 5 cars (3 red, 1 blue, 1 yellow) facing one (given) side:
5C3*2=20
There are two sides -> multiply by 2

20*2 = 40 -> C
Intern
Joined: 30 Jul 2009
Posts: 18

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08 Sep 2009, 14:00
maratikus wrote:
milind1979 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

number of ways to arrange 5 cars (3 red, 1 blue, 1 yellow) facing one (given) side:
5C3*2=20
There are two sides -> multiply by 2

20*2 = 40 -> C

totally ignored that they could all be reversed in.
Intern
Joined: 28 Jan 2010
Posts: 29

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01 Mar 2010, 08:02
Even I feel its 40. But the OA is 20. OA must be wrong.
Intern
Joined: 18 Feb 2010
Posts: 17

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03 Mar 2010, 06:28
4
2
I did it this way:
Step 1: First place the three red cars in 5C3 ways (as all red cars are similar, so arrangement within these red cars has no meaning.)=5C3 = 10
Step 2: The remaining two places have to be filled by two cars (1 blue car and 1 yellow car) = 2! (as these cars can permute amongst themselves, blue is different then yellow)

Step 3: So the answer is 5C3*2! = 10*2 = 20

@maratikus: direction of cars is fixed.

Hope it helps!
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Manager
Joined: 29 Dec 2009
Posts: 66
Location: india

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03 Mar 2010, 07:36
ans shld be 20
5!/3!=20 and v dont display a car facing back side
Intern
Joined: 28 Jan 2010
Posts: 29

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Updated on: 03 Mar 2010, 08:23
1
maratikus wrote:
milind1979 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

number of ways to arrange 5 cars (3 red, 1 blue, 1 yellow) facing one (given) side:
5C3*2=20
There are two sides -> multiply by 2

20*2 = 40 -> C

On second thought, I think the answer is 20.

Also if the direction is to considered, it cannot be limited to just 2. There are infinite number of directions in which the cars can be arranged in.
[For visualizing this, think of the parking space mentioned in question to be a circular one instead of a rectangular one.]

Originally posted by rohityes on 03 Mar 2010, 07:40.
Last edited by rohityes on 03 Mar 2010, 08:23, edited 1 time in total.
Intern
Joined: 28 Jan 2010
Posts: 29

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03 Mar 2010, 07:46
jatt86 wrote:
v dont display a car facing back side

That would be a dangerous assumption to make if not mentioned in the question.

The cars are all arranged to face one direction... thats all we know from the question. The direction can be front, back, north, south, east, west, north-east, south-west etc...
Intern
Joined: 23 Jan 2010
Posts: 28
Schools: Kellogg, Booth, Harvard, Wharton, Stanford
WE 1: Product Strategy
WE 2: Operations
WE 3: Entrepreneurship

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03 Mar 2010, 16:11
1
Why would it not be 20 without the direction.

5!
3!2!

= 10 (we have chosen 3 red cars out of total 5)

10*2 (multiply by two as 1 blue and 1 yellow can be arranged in two ways) = 20
Manager
Joined: 30 Jun 2004
Posts: 147
Location: Singapore

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04 Mar 2010, 00:41
What is the source of the question? And is there any explanation given along with OA?
Manager
Joined: 25 Nov 2011
Posts: 190
Location: India
Concentration: Technology, General Management
GPA: 3.95
WE: Information Technology (Computer Software)

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24 Feb 2012, 05:56
1
rohityes wrote:
maratikus wrote:
milind1979 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

number of ways to arrange 5 cars (3 red, 1 blue, 1 yellow) facing one (given) side:
5C3*2=20
There are two sides -> multiply by 2

20*2 = 40 -> C

On second thought, I think the answer is 20.

Also if the direction is to considered, it cannot be limited to just 2. There are infinite number of directions in which the cars can be arranged in.
[For visualizing this, think of the parking space mentioned in question to be a circular one instead of a rectangular one.]

I totally agree that there could be infinite no. of directions and hence we should not bring this into consideration unless one of the options is infinite.

But, I do not agree to visualize in a circular fashion:
--> in circular way there is no concept of 'direction'
--> when items are arranged in circular fashion, their nCr or nPr values change from that of when items are arranged in a row fashion.
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Current Student
Joined: 26 Jan 2016
Posts: 103
Location: United States
GPA: 3.37
Re: There are 5 cars to be displayed in 5 parking spaces with  [#permalink]

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03 Sep 2016, 14:26
Since the red cars are identical I understand the total number of choices isn't going to be 5! but is the method to divide by 3! because that is the number of duplicate values?
Current Student
Joined: 26 Jan 2016
Posts: 103
Location: United States
GPA: 3.37
Re: There are 5 cars to be displayed in 5 parking spaces with  [#permalink]

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03 Sep 2016, 14:30
Sorry just to add on. My understanding of the method from___ pick___ i.e. we have 20 people how many teams of 5 can we pick is 20!/5!15!

But here I'm confused why we are doing 5!/3!
Is the method to use this when we are making unique groups with a duplicate value--divide by the duplicate value?
Intern
Joined: 06 Jan 2016
Posts: 3
Location: India
Re: There are 5 cars to be displayed in 5 parking spaces with  [#permalink]

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04 Sep 2016, 01:32
in arrangement situation, whenever there is any identical object use this formula
(total object)!/(identical objects)!

in this prob :
5!/3! = 20
Intern
Joined: 10 Jul 2016
Posts: 1
Re: There are 5 cars to be displayed in 5 parking spaces with  [#permalink]

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04 Sep 2016, 02:12
Can someone explain that why have we not considered the situation when 2 red cars are together. Say
R1 R2 Y R3 B is also same as R2 R1 Y R3B...

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Non-Human User
Joined: 09 Sep 2013
Posts: 9162
Re: There are 5 cars to be displayed in 5 parking spaces with  [#permalink]

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28 Oct 2018, 04:28
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Re: There are 5 cars to be displayed in 5 parking spaces with &nbs [#permalink] 28 Oct 2018, 04:28
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