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There are 5 cars to be displayed in 5 parking spaces with

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There are 5 cars to be displayed in 5 parking spaces with [#permalink] New post 07 Aug 2006, 09:35
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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
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 [#permalink] New post 07 Aug 2006, 10:41
(A) 20

Given 5 cars, they can be arranged in 5! ways. However, three are the same color. To eliminate the rearrangement of the three

# of ways = 5!/3! = 20
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 [#permalink] New post 07 Aug 2006, 11:30
Hi guys its me again , i hope am helping , but i think we should put into account the overall directions of the cars in the parking the question is saying facing the same direction.

so the overall display could make difference??????
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 [#permalink] New post 07 Aug 2006, 12:45
I think the statement about direction was provided just so that people don't try to count cases where cars are parked backward, forward, or sideways as different options...

Anyway, I arrived at the same answer from a slightly different angle:

There 5 empty parking slots.

There are 5 options to choose a slot for a Blue car. After the Blue car is parked, we have 4 options to park a Yellow car. The remaining tree slots are taken by Red cars.

So, total number of ways = 5*4 = 20
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Re: PS - Differents Cars arrangements [#permalink] New post 07 Aug 2006, 12:47
The problem clearly states that all cars are to be displayed in the same direction. So no need to consider the direction.

Out of curiousity, what would the answer be if direction is to be considered.

20x2 = 40??



TOUGH GUY wrote:
There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction.
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Re: PS - Differents Cars arrangements [#permalink] New post 07 Aug 2006, 12:58
haas_mba07 wrote:
The problem clearly states that all cars are to be displayed in the same direction. So no need to consider the direction.

Out of curiousity, what would the answer be if direction is to be considered.

20x2 = 40??



TOUGH GUY wrote:
There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction.


Then the answer would be

20 * Infinite = Infinite
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 [#permalink] New post 07 Aug 2006, 13:02
Interesting.. why is it infinite?

Are you including all possible orientations? Then yes it would be infinite... (all the single degree variations of all the cars).


If the options were narrow... say the front of the car can point either north or south only?

Then is it 40?
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 [#permalink] New post 07 Aug 2006, 13:32
In a parking spot in US you can park a car only in 2 possible direction ...ie in the direction of the parking spot..
hence the answer has to be divisible by 2...that eliminates 2 solutions....
I will look more into the combinations and write back
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 [#permalink] New post 07 Aug 2006, 13:51
winsome wrote:
In a parking spot in US you can park a car only in 2 possible direction ...ie in the direction of the parking spot..
hence the answer has to be divisible by 2...that eliminates 2 solutions....
I will look more into the combinations and write back

Who says that the cars are getting displayed in US. :wink: :wink:

Haas,

Excellent question.
I think this will be 20 * 2^5 = 640
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 [#permalink] New post 07 Aug 2006, 14:00
Thanks PS...

It does make sense now...

With 5 cars each one can have 2 orientations... giving 2^5 total options (much like 5 bit logic... 32 possible values)

Now each car (or bit) can have a different position, giving a total of 20 ways of arranging each car.

Total = 20x32 = 640.

Example:
Arranging 2 cars with orientation:
2x2^2 = 8 ways

{ab, ba, a'b, b'a, ab', ba', a'b', b'a'}



ps_dahiya wrote:
winsome wrote:
In a parking spot in US you can park a car only in 2 possible direction ...ie in the direction of the parking spot..
hence the answer has to be divisible by 2...that eliminates 2 solutions....
I will look more into the combinations and write back

Who says that the cars are getting displayed in US. :wink: :wink:

Haas,

Excellent question.
I think this will be 20 * 2^5 = 640
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 [#permalink] New post 07 Aug 2006, 14:46
Folks , i wonder if you can help me with this one , I can't get through it .

It was posted before ( 2003 on the forum)But it is not explained

Ben, Ann, and Ken are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections:

1. How many contain neither Ann nor Ben nor Ken?
2. How many contain Ann but neither Ben nor Ken?
3. How many contain all named persons?
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 [#permalink] New post 07 Aug 2006, 14:53
yezz wrote:
Ben, Ann, and Ken are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections:

1. How many contain neither Ann nor Ben nor Ken?
2. How many contain Ann but neither Ben nor Ken?
3. How many contain all named persons?


1. If all 3 are excluded, then we need to pick 7-3 = 4 out of 4 = 1 way.

2. If 2 excluded, and must have Ann, then first position is taken by Ann; The remaining three need to be filled out of 7-2=5.
Number of ways = 5x4x3/3x2 = 10 ways.

3. All three included; First three positions are taken. Need to fill last one from 7-3=4 people;
Number of ways = 4
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 [#permalink] New post 07 Aug 2006, 15:08
Thanx for the explanation..........

And sure it is time for me to sleep :lol:

See you tomorrow.....take care
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 [#permalink] New post 07 Aug 2006, 15:36
haas,

2. If 2 excluded, and must have Ann, then first position is taken by Ann; The remaining three need to be filled out of 7-2=5.
Number of ways = 5x4x3/3x2 = 10 ways.


Since Ann is in and Ben & Ken are out, should'n we select the remaing 3 semifinalists out of 4 remaining non-named contestants?

So, 3C4 = 4 ???
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 [#permalink] New post 07 Aug 2006, 15:47
Yes... I didn't exclude the "already included Ann" from the count.

Sorry...

So the correct solution would be if 2 excluded and 1 already included we need to pick remaining 3 out of 4 = 4x3x2/3x2 = 4 ways.

Thanks v1rok!!

Sorry yezz, should've checked my work before posting...

v1rok wrote:
haas,

2. If 2 excluded, and must have Ann, then first position is taken by Ann; The remaining three need to be filled out of 7-2=5.
Number of ways = 5x4x3/3x2 = 10 ways.


Since Ann is in and Ben & Ken are out, should'n we select the remaing 3 semifinalists out of 4 remaining non-named contestants?

So, 3C4 = 4 ???
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 [#permalink] New post 08 Aug 2006, 00:37
Haas, Thanks for the help ....you gave me the direction and i should check my self for the valididty.

And yes i agree with the 4 very valid.

Thanks guys .. i just couldnt sleep before finding the answer.
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 [#permalink] New post 08 Aug 2006, 07:02
Guys,

OA is 20:

we have 5 differents options to park a yellow

we do have 4 differents options to park a blue

and the three available spots are available for the three blue one's.

Answer: 5 * 4 = 20
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  [#permalink] 08 Aug 2006, 07:02
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