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plaza202
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counting methods 11 Nov 2007, 21:23
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there are 5 cars to be displayed in 5 paring spaces with all the cars facing the same direction. of the 5 cars, 3 are red, 1 blue and 1 yellow. if the cars are identical except for the color, how many different display arrangements of the 5 cars are possible?

20
25
40
60
125...



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counting methods 11 Nov 2007, 21:37
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Option A is the answer

Ignore the red cars. We will concentrate on the yellow and the blue car.
There are 5 possible positions for the yellow car. For each such position of the yellow car, there are 4 possible positions for the blue car. Therefore the answer is 5*4 =20.
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counting methods 12 Nov 2007, 01:32
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jbs
Option A is the answer

Ignore the red cars. We will concentrate on the yellow and the blue car.
There are 5 possible positions for the yellow car. For each such position of the yellow car, there are 4 possible positions for the blue car. Therefore the answer is 5*4 =20.
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Can you pls. explain why you ignore the number of red cars?
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counting methods 12 Nov 2007, 04:14
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GK_Gmat
jbs
Option A is the answer

Ignore the red cars. We will concentrate on the yellow and the blue car.
There are 5 possible positions for the yellow car. For each such position of the yellow car, there are 4 possible positions for the blue car. Therefore the answer is 5*4 =20.

Can you pls. explain why you ignore the number of red cars?
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They say that the cars are identical in every way except for the colour.

Therefore, once you have decided on the positions of the yellow and the blue car, the 3 red cars can be arranged in any manner among themselves to give the same display arrangement.

Here's an example. For the sake of understanding, assume that the 3 red cars are r1,r2 and r3.

here's a possible display arrangement

Y B r1 r2 r3

now note that the above display arrangement will look exactly the same as

Y B r2 r3 r1

My understanding of the question tells me that the exact position of a particular car is not important but the display arrangement is.

Therefore, as we seek to find the number of different display arrangements, we can ignore the similar red cars.

Also, refer to GMAT TIGER's post for a more conceptual approach.
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counting methods 12 Nov 2007, 12:36
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jbs
GK_Gmat
jbs
Option A is the answer

Ignore the red cars. We will concentrate on the yellow and the blue car.
There are 5 possible positions for the yellow car. For each such position of the yellow car, there are 4 possible positions for the blue car. Therefore the answer is 5*4 =20.

Can you pls. explain why you ignore the number of red cars?

They say that the cars are identical in every way except for the colour.

Therefore, once you have decided on the positions of the yellow and the blue car, the 3 red cars can be arranged in any manner among themselves to give the same display arrangement.

Here's an example. For the sake of understanding, assume that the 3 red cars are r1,r2 and r3.

here's a possible display arrangement

Y B r1 r2 r3

now note that the above display arrangement will look exactly the same as

Y B r2 r3 r1

My understanding of the question tells me that the exact position of a particular car is not important but the display arrangement is.

Therefore, as we seek to find the number of different display arrangements, we can ignore the similar red cars.

Also, refer to GMAT TIGER's post for a more conceptual approach.
Show more


perfect. in repetitive case,

req possibilities = total possibilities/repetitive possibilities
req possibilities = 5!/3!


because r1 r2 r3 and r3 r2 r1 are same and counted as 1 possibility. so we need to divide 5! by 3!.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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