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

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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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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 $$\frac{5!}{3!}=20$$.

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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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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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).

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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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Hi

Can anyone please explain, why the answer here is not 40. (5!/3!)*2
In a parking slot all cars can face 2 directions right? and these 2 directions will comprise of 2 different arrangements.
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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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vinayakvaish wrote:
Hi

Can anyone please explain, why the answer here is not 40. (5!/3!)*2
In a parking slot all cars can face 2 directions right? and these 2 directions will comprise of 2 different arrangements.

I second 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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RiseAndShinee wrote:
vinayakvaish wrote:
Hi

Can anyone please explain, why the answer here is not 40. (5!/3!)*2
In a parking slot all cars can face 2 directions right? and these 2 directions will comprise of 2 different arrangements.

I second this.

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

I can see how someone could read the above and conclude that we must consider two different scenarios:
1) all of the cars facing in one direction
2) all of the cars facing in the opposite direction

However, if we go down this rabbit hole, we must recognize that there are infinitely many directions the cars could be facing. For example they could all be facing north. Or they could all be facing 1° east of north. Or they could all be facing 1.00253° east of north. etc.
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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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RiseAndShinee wrote:
vinayakvaish wrote:
Hi

Can anyone please explain, why the answer here is not 40. (5!/3!)*2
In a parking slot all cars can face 2 directions right? and these 2 directions will comprise of 2 different arrangements.

I second this.

Note that when we make people stand in a line, we do not worry about how they are facing and the assumption is that they all are facing in the same direction. The arrangements are relative to each other, not relative to the environment. Here, the questions mentions that all cars are facing the same direction. Whether they all are facing east or all are facing west doesn't matter because relative to each other, the arrangements are the same.

Anyway, don't worry. Before questions go live, they are experimental for a fair bit of time to ensure that such issues are rectified. Say if 50% people with scores 650+ answer 20 and 50% answer 40, they know that the statements in the question are being evaluated differently by different people.
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars [#permalink]
It's a parking lot/space. Logic dictates there are parking spaces/lines defined, thus 2 directions a car could face within those lines. I know 20 is the 'official' answer, but the fact that GMAC included this bit of information in the question stem makes it very confusing, because now those of us that read the question and figured there'd be a x2 multiplier at the end of the arrangement would end up with the 'wrong' answer of 40. Agree to disagree.
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars [#permalink]
juxmba wrote:
It's a parking lot/space. Logic dictates there are parking spaces/lines defined, thus 2 directions a car could face within those lines. I know 20 is the 'official' answer, but the fact that GMAC included this bit of information in the question stem makes it very confusing, because now those of us that read the question and figured there'd be a x2 multiplier at the end of the arrangement would end up with the 'wrong' answer of 40. Agree to disagree.

Completely understand your frustration. But it is also given that the cars are facing the same direction - telling you that the direction should not be accounted for.
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There are 5 cars to be displayed in 5 parking spaces with all the cars [#permalink]
AnirudhaS wrote:
juxmba wrote:
It's a parking lot/space. Logic dictates there are parking spaces/lines defined, thus 2 directions a car could face within those lines. I know 20 is the 'official' answer, but the fact that GMAC included this bit of information in the question stem makes it very confusing, because now those of us that read the question and figured there'd be a x2 multiplier at the end of the arrangement would end up with the 'wrong' answer of 40. Agree to disagree.

Completely understand your frustration. But it is also given that the cars are facing the same direction - telling you that the direction should not be accounted for.

I have read the comments "defending" this question, but we have to agree that this question is poorly written.

We get three restrictions:
1) The cars are identical except for color. -> Ok, we get that we have to use $$5!/3!$$...
2) They are to be displayed in 5 parking spaces. -> Ok, so they have to be parked within these parking spaces.
3) The cars face the same direction. -> This can certainly be interpreted in two ways: All cars must face one specific direction or the cars must face the same direction relative to each other. (Either way it is important that this restriction is given, so that test-takers don't account for the different instances of exactly one car facing the opposite direction, or two cars, three cars etc.)

The main question is: How many different display arrangements of the 5 cars are possible?
So we have 20 different possible orders for the cars and they can either be arranged all facing one way or the other way... So 40 different solutions.

GMAC should have written the question as: How many different orders of the 5 cars are possible?.
If that had been the question, changing the direction of the cars would simply be a $$180^{\circ}$$ turn of some order already accounted for.

I feel this question is punishing the test-takers who implement critical thinking. That is just my opinion.
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars [#permalink]
BrentGMATPrepNow wrote:
RiseAndShinee wrote:
vinayakvaish wrote:
Hi

Can anyone please explain, why the answer here is not 40. (5!/3!)*2
In a parking slot all cars can face 2 directions right? and these 2 directions will comprise of 2 different arrangements.

I second this.

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

I can see how someone could read the above and conclude that we must consider two different scenarios:
1) all of the cars facing in one direction

2) all of the cars facing in the opposite direction

However, if we go down this rabbit hole, we must recognize that there are infinitely many directions the cars could be facing. For example they could all be facing north. Or they could all be facing 1° east of north. Or they could all be facing 1.00253° east of north. etc.

BrentGMATPrepNow

I am a bit confused as to how you would proceed if you do use the permutation formula. I came across two versions of the permutation formula (one with repetitions) and one without repetitions. What does the one with repetitions looks like? You use take n!/repetitions!*non-repetitions! Are there repetition and non-repetition formula for combinations?

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

Cheers,
Brent

BrentGMATPrepNow

I was struggling with this one and saw "arrangements" so thought this is a permutation problem.

Key words I have learned to be associated with permutations:
-Arrange
-Line up
-Order

Key words I have learned to be associated with combinations:
-Choose
-Select
-Pick

I am assuming now that we cannot always rely on these key words for guidance in figuring out if a problem is a permutation problem vs. a combinatorics problem.

Many thanks
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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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woohoo921 wrote:
BrentGMATPrepNow wrote:
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

Cheers,
Brent

BrentGMATPrepNow

I was struggling with this one and saw "arrangements" so thought this is a permutation problem.

Key words I have learned to be associated with permutations:
-Arrange
-Line up
-Order

Key words I have learned to be associated with combinations:
-Choose
-Select
-Pick

I am assuming now that we cannot always rely on these key words for guidance in figuring out if a problem is a permutation problem vs. a combinatorics problem.

Many thanks

This question is pretty much a permutation question, EXCEPT the denominator in the MISSISSIPPI formula helps deal with the identical objects to be arranged.
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars [#permalink]
BrentGMATPrepNow
Thank you! Out of curiosity, if you were using the permutation formula, I am confused as to what you would plug in...
5!/(5-2)!

Is this correct above? I see that the other answers on the form have 5!/3!, but I am confused as to why we would plug in 2 as the "r" in the formula.

Thanks again
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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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woohoo921 wrote:
BrentGMATPrepNow
Thank you! Out of curiosity, if you were using the permutation formula, I am confused as to what you would plug in...
5!/(5-2)!

Is this correct above? I see that the other answers on the form have 5!/3!, but I am confused as to why we would plug in 2 as the "r" in the formula.

Thanks again

We are taking all 5 letters (R, R, R, B and Y) and permuting all 5.
We can do this in 5! ways.
However, the 3 identical R's mean we are counting each arrangement more than once.
In fact, we are counting each arrangement 6 times (since we can arrange 3 R's in 3! ways)
So, the total number of DISTINCT arrangements = 5!/3!

BTW, we don't really need to memorize the Permutation formula for the GMAT. More here: https://www.gmatprepnow.com/articles/pe ... las-forget
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars [#permalink]
Just curious how this would be solved if the cars were allowed to face in different directions?
Re: There are 5 cars to be displayed in 5 parking spaces with all the cars [#permalink]
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