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

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swerve: Legendary Member; Posts: 2535; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Wed Oct 10, 2018 9:34 am

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

The OA is A

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Source: Beat The GMAT — Problem Solving | Wed Oct 10, 2018 9:34 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Oct 10, 2018 9:59 am

swerve 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

Let R, R, R, B, Y represent the cars (by their colors)
Notice that the three R's are identical.
So, the question becomes In how many different ways can we arrange the letters R, R, R, B and Y?

----------------ASIDE------------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
-------------ONTO THE QUESTION!------------------------------

We have R, R, R, B and Y:
There are 5 letters in total
There are 3 identical R's
So, the total number of possible arrangements = 5!/(3!)
= (5)(4)(3)(2)(1)/(3)(2)(1)
= 20

Answer: A

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Oct 10, 2018 10:16 am

swerve 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

Alternate approach:

The 3 red cars must occupy a combination of 3 spaces.
From 5 spaces, the number of ways to choose a combination of 3 for the red cars = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of options for the blue car = 2. (Either of the 2 remaining spaces.)
Number of options for the yellow car = 1. (Only 1 space left.)
To combine the options above, we multiply:
10*2*1 = 20.

The correct answer is A.

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