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counting methods [#permalink] New post 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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 [#permalink] New post 11 Nov 2007, 21:37
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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 [#permalink] New post 12 Nov 2007, 01:32
jbs wrote:
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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 [#permalink] New post 12 Nov 2007, 04:14
GK_Gmat wrote:
jbs wrote:
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.
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 [#permalink] New post 12 Nov 2007, 12:36
jbs wrote:
GK_Gmat wrote:
jbs wrote:
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.


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!.
  [#permalink] 12 Nov 2007, 12:36
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