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by crackgmat007 » Wed May 06, 2009 9:39 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

Is C not the correct answer? I got 20 as the number of arrangements if the cars are displayed in one direction. But if we reverse the direction of all cars, the total arrangements will be 20*2=40. Did I assume beyond what is stated in the problem?
OA – A
Last edited by crackgmat007 on Sun May 10, 2009 9:31 pm, edited 1 time in total.
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by DanaJ » Wed May 06, 2009 10:15 am
You are right, 20 is the right answer. You are not supposed to consider the cars facing in different ways, since that's exactly why you have that statement in there:

all the cars facing the same direction

That basically tells you that they share the direction and that you shouldn't worry about it. Just "worry" about the colors!
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by apple100 » Sun May 10, 2009 6:24 am
DanaJ wrote:You are right, 20 is the right answer. You are not supposed to consider the cars facing in different ways, since that's exactly why you have that statement in there:

all the cars facing the same direction

That basically tells you that they share the direction and that you shouldn't worry about it. Just "worry" about the colors!
How did you get 20?
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by Svedankae » Sun May 10, 2009 12:45 pm
It's 5! / 3!*1!*1!
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by apple100 » Sun May 10, 2009 2:37 pm
Svedankae wrote:It's 5! / 3!*1!*1!

what is the formula for these questions?
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by crackgmat007 » Sun May 10, 2009 9:35 pm
Factorial of total # of cars/Factorial of cars that are identical.

Since there are 5 cars, factorial will be 5!
3 cars are red in color, hence 3!

5!/3! gives us 20. Hope this helps
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