Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

17 Mar 2009, 20:21

1

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

73% (01:53) correct
27% (00:29) wrong based on 104 sessions

HideShow timer Statistics

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

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.

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)

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.

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...

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.
_________________

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

Show Tags

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?

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

Show Tags

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?