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There are 5 cars to be displayed in 5 parking spaces with [#permalink]
17 Mar 2009, 20:21

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Question Stats:

0% (00:00) correct
100% (00:18) wrong based on 11 sessions

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

Re: Permutation + combination [#permalink]
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!

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.

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

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.

Re: Permutation + combination [#permalink]
03 Mar 2010, 06:28

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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)

Re: Permutation + combination [#permalink]
03 Mar 2010, 07:40

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, 08:23, edited 1 time in total.

Re: Permutation + combination [#permalink]
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...

Re: Permutation + combination [#permalink]
24 Feb 2012, 05:56

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. _________________

------------------------- -Aravind Chembeti

gmatclubot

Re: Permutation + combination
[#permalink]
24 Feb 2012, 05:56