ashg84 wrote:There are 5 cars to be displayed in 5 parking spaces with all the cars facing direction. Of the 5 cars, 3 are red, 1 is blue, and 1 is yellow. If the cars are identical except for the color, how many different display arrangement of the 5 cars are possible?
a - 20
b - 25
c - 40
d - 60
e - 125
The number of ways to arrange 5 DISTINCT elements = 5! = 120.
But here, the 3 red cars are IDENTICAL.
When identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
To account for the duplicate arrangements, we DIVIDE by the number of ways to ARRANGE the identical elements.
The number of ways to arrange the 3 red cars = 3!.
Thus:
5!/3! = 20.
The correct answer is
A.
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