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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There are 5 cars to be displayed in 5 parking spaces with
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Let R, R, R, B, Y represent the cars (by their colors)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
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
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Alternate approach: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
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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