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

[M7MBA]

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

Answer: A

Source: Official Guide

[Brent@GMATPrepNow]

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

Answer: A

Source: Official Guide

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

[deloitte247]

Total number of cars = 5
Total number of identical cars = 3 red cars
Total number of possible display arrangement = 5!/3!
$$\frac{5!}{3!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1}$$
$$=\ 20$$
$$Answer\ =\ A$$

[Scott@TargetTestPrep]

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

Answer: A

Source: Official Guide

Solution:
We are given that in order to make orange juice, 1 can of concentrate is used and 3 cans of water. Thus, we can set up the following ratio:
Concentrate : water = x : 3x
We need to determine how many cans of concentrate are needed to prepare 200 6-ounce servings of orange juice, or 200 x 6 = 1,200 ounces.
We can create the following equation to determine the number of ounces of concentrate needed:
x + 3x = 1,200
4x = 1,200
x = 300 ounces
Since there are 12 ounces per can, 300/12 = 25 cans are needed to make 1,200 ounces of orange juice.
Answer: A