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

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

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New post 16 Dec 2009, 21:53
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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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There are 5 cars to be displayed in 5 parking spaces with all the cars  [#permalink]

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New post 17 Dec 2009, 01:19
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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

I think the above logic is correct except one thing:

# of arrangements is \(\frac{5!}{3!}=20\).

Answer: A.

Though I think that the question is quite ambiguous. We are told that "all the cars are facing the SAME direction", not that they all are facing ONE specific direction. That's why I think we should multiply by 2.

A car can be arranged facing TWO directions (within one particular parking space), so we should multiply 20 by 2, \(20*2=40\)

They can be arranged as
-->
-->
-->
-->
-->
20 combinations.

AND
<--
<--
<--
<--
<--
20 combinations.

Stem just says that they can can not be arranged for example like this:
<--
-->
-->
<--
<--

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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars  [#permalink]

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New post 16 Dec 2009, 22:45
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I'm going A.

I think you just think of it as 2 different cars and 5 empty spots, since the rest have to be red anyway. Then there's 5 spaces for the yellow car and four spaces for the blue one for each yellow car space. 4x5 = 20

Probably if I could remember the maths from permutations and combinations you'd end up with something like 5!/3!
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars  [#permalink]

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New post 10 Oct 2018, 10:02
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xcusemeplz2009 wrote:
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


Let R, R, R, B, Y represent the cars (by their colors)
Notice that the three R's are identical.
So, the question becomes In how many different ways can we arrange the letters R, R, R, B and Y?

----------------ASIDE------------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
-------------ONTO THE QUESTION!------------------------------

We have R, R, R, B and Y:
There are 5 letters in total
There are 3 identical R's
So, the total number of possible arrangements = 5!/(3!)
= (5)(4)(3)(2)(1)/(3)(2)(1)
= 20

Answer: A

Cheers,
Brent
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There are 5 cars to be displayed in 5 parking spaces with all the cars  [#permalink]

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New post 10 Oct 2018, 13:22
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xcusemeplz2009 wrote:
There are 5 cars to be displayed in 5 parking spaces with all the cars facing A FIXED 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

\({\text{cars}}\,\,\,\left\{ \begin{gathered}
\,3\,\,{\text{red}} \hfill \\
\,1\,\,{\text{blue}} \hfill \\
\,1\,\,{\text{yellow}} \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\)

\(?\,\,\,:\,\,\,\# \,\,\,{\text{in}}\,\,5\,\,{\text{parking}}\,\,{\text{spaces}}\,\,\,\,\left( {{\text{only}}\,\,{\text{colors}}\,\,{\text{matter}}} \right)\)


\(?\,\,\, = \,\,\,\underbrace {C\left( {5,3} \right)}_{{\text{red}}\,\,{\text{parking}}\,\,{\text{choices}}}\,\,\, \cdot \,\,\,\underbrace {C\left( {2,1} \right)}_{{\text{blue}}\,\,{\text{parking}}\,\,{\text{choices}}}\,\,\,\,\, = \,\,\,\,\,10 \cdot 2\,\,\, = 20\)

(Once chosen the red and blue parking spaces - among 5 and 2 available, respectively -, the yellow car is placed in the remaining parking space left.)

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars  [#permalink]

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New post 13 Oct 2018, 17:28
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xcusemeplz2009 wrote:
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


The cars can be displayed in the following number of ways, using the indistinguishable permutations formula:

(5!)/(3!) = 5 x 4 = 20 ways.

Note that we divided by 3! because the 3 red cars are identical to each other (i.e., they are indistinguishable).

Answer: A
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars  [#permalink]

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New post 14 Oct 2018, 01:32
I went about this a slightly longer way;

3 Red cars, 5 spaces -> 15 possibilities
1 Blue car, 5 spaces -> 5 possibilities
1 Yellow car, 5 spaces -> 5 possibilities

Total number of arrangements: 15+5+5 = 20.
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There are 5 cars to be displayed in 5 parking spaces with all the cars  [#permalink]

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New post 19 Jan 2019, 02:35
swatato wrote:
I went about this a slightly longer way;

3 Red cars, 5 spaces -> 15 possibilities
1 Blue car, 5 spaces -> 5 possibilities
1 Yellow car, 5 spaces -> 5 possibilities

Total number of arrangements: 15+5+5 = 20.


no .... 15+5+5 <> 20 :roll:
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There are 5 cars to be displayed in 5 parking spaces with all the cars   [#permalink] 19 Jan 2019, 02:35
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