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

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.

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

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


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

(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.

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

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.
Please can some expert explain?
Bunuel VeritasKarishma BrentGMATPrepNow

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.

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.

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.

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

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.

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

Could you do 5 choose 3 for red then (5 choose 1) *2? also yields the answer of 20

­
Yes, if the logic behind these calculations is that 5C3 is the number of ways to choose 3 spaces for the red cars out of 5 (since the red cars are identical, this accounts for the three unordered spaces), and 2! is the number of ways to arrange the blue and yellow cars in the remaining two spaces, then:

5C3 * 2! = 20.

Answer: A.