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

Question

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

(A) 20
(B) 25
(C) 40
(D) 60
(E) 125

PS10309

Bunuel · Accepted Answer

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

(A) 20
(B) 25
(C) 40
(D) 60
(E) 125

This is a simple problem of arranging 3 red, 1 blue and 1 yellow. The number of arrangements is  \frac{5!}{3!}=20 .

Answer: A.

BrentGMATPrepNow · Answer

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

fskilnik · Answer

{	ext{cars}}\,\,\,\left\{ \begin{gathered}
 \,3\,\,{	ext{red}} \hfill \
 \,1\,\,{	ext{blue}} \hfill \
 \,1\,\,{	ext{yellow}} \hfill \ 
\end{gathered} ight.\,\,\,\,\,\,

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

?\,\,\, = \,\,\,\underbrace {C\left( {5,3} ight)}_{{	ext{red}}\,\,{	ext{parking}}\,\,{	ext{choices}}}\,\,\, \cdot \,\,\,\underbrace {C\left( {2,1} ight)}_{{	ext{blue}}\,\,{	ext{parking}}\,\,{	ext{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.

ScottTargetTestPrep · Answer

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

vinayakvaish · Answer

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.

BrentGMATPrepNow · Answer

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.

KarishmaB · Answer

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.

woohoo921 · Answer

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

BrentGMATPrepNow · Answer

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

dd2151 · Answer

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

Bunuel · Answer

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.

mandidaw · Answer

Hi! why is it two different cars? Can't it be seen as 3 different cars cause there are 3 distinct colors?

Bunuel · Answer

It’s because the three red cars are identical, so switching them doesn’t create a new arrangement. Only the positions of the blue and yellow cars matter. Check the discussion above for more.

PGTLrowanhand · Answer

Here's my video solution:  https://www.youtube.com/watch?v=wiqRKGpvHMo

sunshineeee · Answer

I use this approach

5 C 3 + 5 C 1 + 5 C 1 -> refer to red+blue+yellow
= 10 + 5 + 5 
= 20

Pls advise: does it go correctly as well? Thank you

Bunuel · Answer

The final answer is 20, but that method is not correct. The choices are not added because placing the red, blue, and yellow cars are not separate alternatives.

Choose 3 spaces for the red cars, then 1 of the remaining 2 spaces for the blue car:

5C3 * 2C1 = 10 * 2 = 20

The yellow car occupies the last space. Your calculation gives 20 only by coincidence.

For example:

There are 6 cars to be displayed in 6 parking spaces. Of the 6 cars, 4 are red, 1 is blue, and 1 is yellow. The cars are identical except for color. How many different arrangements are possible?

Correct method:

6C4 * 2C1 = 15 * 2 = 30

Or:

6!/4! = 30

Using the incorrect method:

6C4 + 6C1 + 6C1 = 15 + 6 + 6 = 27

So the addition method does not work. In the original question, it gave 20 only by coincidence.

egmat · Answer

Hi sunshineeee,

Good instinct starting with  5C3  for the red cars: choosing  3  spaces out of  5  for the identical reds is exactly right, and that's your  10 .

The problem is what happens after that. Your method  adds  the pieces ( 10  +  5  +  5 ), and it also treats blue and yellow as each having  5  spaces to choose from. Both of those are off - it just happens to land on  20  by coincidence.

Two fixes:

- Once the  3  red spaces are taken, only  2  spaces are left, not  5 . So blue has  2  choices, and yellow takes the  1  remaining space. Not  5C1  =  5  each.
- When you make choice A  and then  choice B, you  multiply , you don't add. Adding is only for "either this case  or  that case" (mutually exclusive scenarios). Placing red  and  blue  and  yellow is one combined process, so it's a product:

5C3  ×  2  ×  1  =  10  ×  2  =  20

That's the same  20  the other solutions got - but through the correct "AND = multiply" logic.

Why the coincidence is dangerous

Adding gave you the right number here by luck. Shrink the problem to see it break:

Arrange  2  red,  1  blue in  3  spaces. List them by hand:  RRB, RBR, BRR  - that's  3  arrangements.

- Correct method:  3C2  (choose red spaces) ×  1  (blue takes the last) =  3 . ✓
- Your addition method:  3C2  +  3C1  =  3  +  3  =  6 . ✗

The addition approach doubles the true count. So the rule to carry forward:  combined steps multiply; only separate either/or cases add  - and always count spaces as they get used up.

Answer: A

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

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