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There are 5 cars to be displayed in 5 parking spaces with [#permalink]
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17 Mar 2009, 21:21
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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 == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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Re: Permutation + combination [#permalink]
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17 Mar 2009, 21:35
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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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Re: Permutation + combination [#permalink]
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24 Mar 2009, 16:29
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!



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Re: Permutation + combination [#permalink]
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08 Sep 2009, 14:15
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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.



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Re: Permutation + combination [#permalink]
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08 Sep 2009, 14:50
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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



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Re: Permutation + combination [#permalink]
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08 Sep 2009, 15:00
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.



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Re: Permutation + combination [#permalink]
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01 Mar 2010, 09:02
Even I feel its 40. But the OA is 20. OA must be wrong.



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Re: Permutation + combination [#permalink]
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03 Mar 2010, 07:28
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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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Re: Permutation + combination [#permalink]
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03 Mar 2010, 08:36
ans shld be 20 5!/3!=20 and v dont display a car facing back side



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Re: Permutation + combination [#permalink]
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03 Mar 2010, 08:40
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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.]
Last edited by rohityes on 03 Mar 2010, 09:23, edited 1 time in total.



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Re: Permutation + combination [#permalink]
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03 Mar 2010, 08:46
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, northeast, southwest etc...



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Re: Permutation + combination [#permalink]
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03 Mar 2010, 17:11
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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



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Re: Permutation + combination [#permalink]
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04 Mar 2010, 01:41
What is the source of the question? And is there any explanation given along with OA?



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Re: Permutation + combination [#permalink]
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24 Feb 2012, 06:56
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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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Re: There are 5 cars to be displayed in 5 parking spaces with [#permalink]
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03 Sep 2016, 15:26
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?



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Re: There are 5 cars to be displayed in 5 parking spaces with [#permalink]
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03 Sep 2016, 15:30
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 valuedivide by the duplicate value?



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Re: There are 5 cars to be displayed in 5 parking spaces with [#permalink]
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04 Sep 2016, 02:32
in arrangement situation, whenever there is any identical object use this formula (total object)!/(identical objects)!
in this prob : 5!/3! = 20



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Re: There are 5 cars to be displayed in 5 parking spaces with [#permalink]
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04 Sep 2016, 03:12
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... Sent from my Lenovo A7010a48 using GMAT Club Forum mobile app



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