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There are 5 cars to be displayed in 5 parking spaces with [#permalink]
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?

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??????

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.

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 _________________

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.

Haas,

Excellent question.
I think this will be 20 * 2^5 = 640 _________________

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.

Haas,

Excellent question. I think this will be 20 * 2^5 = 640

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

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?

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?

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...