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The Carson family will purchase three used cars. There are

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The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

A. 24
B. 32
C. 48
D. 60
E. 192

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Originally posted by enigma123 on 11 Mar 2012, 01:03.
Last edited by Bunuel on 13 Jul 2013, 06:51, edited 1 time in total.
Added the OA.
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New post 11 Mar 2012, 01:18
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enigma123 wrote:
The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

A. 24
B. 32
C. 48
D. 60
E. 192

Any idea how to solve this guys? I don't have an OA unfortunately.


\(C^3_4*2^3=4*8=32\), \(C^3_4\) selecting 3 different colors from 4 and multiplying by 2*2*2=2^3 since each color car has two options: model A or model B.

Answer: B.

Similar question to practice:
if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html
if-a-committee-of-3-people-is-to-be-selected-from-among-98533.html

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New post 09 May 2012, 08:26
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ashish8 wrote:
calreg11 wrote:
Bunuel, I don't understand the logic behind why you grouped it like that.
in MGMAT they said to use the slot method, first car has 8 choices, second has 6, third has 6. to account for overcounting we divide by the factorial 3!. same number but i understand that logic a little better but i am inclined to use your method on the test.
Can you explain?


Bunuel, Calgreg, if I follow the MGMAT method I don't understand what over counting they are talking about. I understand the 8 for the first choice, 6 for the second, and 4 for the third, but why do the division afterwards?


The method of selecting 8 for the first choice, 6 for the second and 4 for the third is what we call 'basic counting principle'. When you do 8*6*4, you are effectively selecting and ARRANGING the cars: you say 'The FIRST car is selected in 8 ways, the SECOND car in 6 ways etc'. But you don't have a first second third car. You only have a group of 3 cars. So to un-arrange (so to say), you need to divide by 3!

I have explained this concept in this post: http://www.veritasprep.com/blog/2011/11 ... binations/
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New post 26 Mar 2012, 23:41
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calreg11 wrote:
Bunuel, I don't understand the logic behind why you grouped it like that.
in MGMAT they said to use the slot method, first car has 8 choices, second has 6, third has 6. to account for overcounting we divide by the factorial 3!. same number but i understand that logic a little better but i am inclined to use your method on the test.
Can you explain?


We are selecting 3 different color cars out of 4 possible colors. In how many ways it can be done? \(C^3_4=4\), selecting 3 out of 4.

Next, there are 2 models of each selected car of a certain color available, hence each selected car has 2 options: Model A or Model B. Since there are 3 selected cars then total ways is 2*2*2.

Grand total 4*2^3=32.

Check the links in my previous post for similar questions.

Hope it helps.
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New post 26 Dec 2012, 23:33
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eaakbari wrote:
Bunuel, Karishma,

You gotta help me out here, m getting shaky on all my Combinatorics concepts.

I approached the problem as

\(8C8 * 7C6 * 4C1\)

What exactly m I doing wrong?


I have no idea how you got 8C8, 7C6 and 4C1.

I could have understood 8C1*6C1*4C1 (I have explained why this doesn't work in the post above)

The best way to solve it is by first selecting 3 colors out of the given 4 in 4C3 ways. (say, you got black, red and green)
Now for each color, you have 2 choices - model A or B
So you select a model in 2 ways.
No of ways of selecting the 3 cars = 4C3 * 2 * 2 * 2 = 32
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New post 27 Dec 2012, 00:28
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eaakbari wrote:

Well, my logic was

The number of ways of selecting r objects from n different objects is nCr.

We have 8 options and can select either of the 8 as the first - 8C8

One car is gone and one becomes a forbidden choice so we have 7 options and need to select either of the 6 as the first - 7C6

Two cars are gone and two becomes a forbidden choice so we have 6 options and need to select either of the 4 as the first - 6C4

I reckon that my understanding of Combinatorics logic is fundamentally flawed. Appreciate if you could tell me where and what


Yes, I think you need to go through a PnC book from scratch. You can check out the Veritas Combinatorics and Probability book at amazon.

"The number of ways of selecting r objects from n different objects is nCr" - This is correct. You need to apply it correctly now.

As for this question,

We have 8 options and can select either of the 8 as the first - 8C8
--- We have 8 options and we need to select 1 out of those 8. So n = 8 and r = 1. This gives us 8C1 ways. (Say you selected BlackA)

One car is gone and one becomes a forbidden choice so we have 7 options and need to select either of the 6 as the first - 7C6
--- Now one car is gone and we cannot select another one. Now we have 6 cars to choose from and we again have to select 1. So n = 6 and r = 1. We get 6C1 ways (Say you selected GreenB)

Two cars are gone and two becomes a forbidden choice so we have 6 options and need to select either of the 4 as the first - 6C4
--- Now 2 cars are gone and 2 are forbidden. You can choose out of only 4 cars and you have to select only 1. Again, n = 4 and r = 1 so we get 4C1 (Say you selected RedA)

You select 3 cars in 8C1 * 6C1 * 4C1 ways but there is a flaw here. You selected BlackA, GreenB and RedA.
In another case, you could have selected RedA first, then BlackA and then GreenB. This combination is the same as the previous one but we are counting it separately. The point is we have arranged the cars as first, second and third which we should not do since it is a group.

As I said before, you might want to start with fundamentals from a book.
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New post 03 Jun 2014, 20:21
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VeritasPrepKarishma wrote:

The best way to solve it is by first selecting 3 colors out of the given 4 in 4C3 ways. (say, you got black, red and green)
Now for each color, you have 2 choices - model A or B
So you select a model in 2 ways.
No of ways of selecting the 3 cars = 4C3 * 2 * 2 * 2 = 32


Quote:
My doubt was

Can you please explain why we have multiplied 2*2*2 with 4C3.

what i did to solve this was 4C3 * 2C1 where 2C1 are for selection of model . Can you please let me know what i have missed here.

Thanks


What you did is correct but incomplete.

You select 3 colors in 4C3 ways is correct. Say you select blue, black and red.
Now you have 3 different colors but for each color you have 2 models available. So you must select a model for EACH color. You do that in 2C1 ways for EACH color. Say model A for blue, model B for black and model A for red.

Total selection can be made in 4C3 * 2C1 * 2C1 * 2C1 ways
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New post 27 Jun 2014, 22:12
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deya wrote:
Hello,
I have solved the problem in the following way. From the question I came to know that there are two categories of cars of 4 different colors each. So, in total there are 8 cars. We need to select 3 cars of different colors. So, we can do it in two ways. We can select 2 cars from category A and one from B------> C(4,2)* C(2,1) = 12 ----> C(2,1) for B as we have already selected two colors from A so we are left with the rest 2 colors and we have selected 1 from them.

Secondly, we can select 1 car from A and 2 cars from B-----> C(4,1) * C(3,2) = 12

Therefore, total ways of combinations = 12+12 = 24. But, unfortunately this is not the correct answer. Kindly tell me where I went wrong.

Thank You in advance. :-D


What about selecting all three cars from A or all three cars from B?

All three cars can have the same model. The only constraint is that they need to be of different colors.

Select all from A - In 4C3 = 4 ways
Select all from B - In 4C3 = 4 ways

Total = 24+4+4 = 32 ways
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New post 26 Oct 2015, 21:34
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davesinger786 wrote:
VeritasPrepKarishma wrote:
deya wrote:
Hello,
I have solved the problem in the following way. From the question I came to know that there are two categories of cars of 4 different colors each. So, in total there are 8 cars. We need to select 3 cars of different colors. So, we can do it in two ways. We can select 2 cars from category A and one from B------> C(4,2)* C(2,1) = 12 ----> C(2,1) for B as we have already selected two colors from A so we are left with the rest 2 colors and we have selected 1 from them.

Secondly, we can select 1 car from A and 2 cars from B-----> C(4,1) * C(3,2) = 12

Therefore, total ways of combinations = 12+12 = 24. But, unfortunately this is not the correct answer. Kindly tell me where I went wrong.

Thank You in advance. :-D


What about selecting all three cars from A or all three cars from B?

All three cars can have the same model. The only constraint is that they need to be of different colors.

Select all from A - In 4C3 = 4 ways
Select all from B - In 4C3 = 4 ways

Total = 24+4+4 = 32 ways



Hi Karishma,
Need some help here.

I tried the method given above as well.But I'm confused regarding this.If I'm selecting 2 cars of Model A in 4c2 ways and correspondingly I select I car from Model B in 2c1 ways(2 colors removed because already selected in A) then why are not taking the cases other way around i.e 2 cars of Model B in 4c2 and 1 car of B in 2c1.After all,they are all different cars.Won't that be added to the combinations as well. In that case, the number of combinations would become (12+12+4)*2=56 ways..Kindly put your thoughts here.


Right, so 4C2 * 2C1 = 12 (2 cars of model A and 1 car of model B)
From where do we get the other 12? It represents the case where we select 2 cars of model B and 1 car of model A. So these are already accounted for.

Next, you select all from A - In 4C3 = 4 ways
and then select all from B - In 4C3 = 4 ways

Total = 12+12+4+4 = 32 ways
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RAHKARP27071989 wrote:


Hi VeritasPrepKarishma,

In this question I am just doing 8C1*6C1*4C1

As you said, by doing this we are selecting and then arranging the cars. Therefore this approach is wrong.
But here I am applying Combination, which ONLY select the things.

If we would have done 8P1*6P1*4P1--> it means we are first selecting and then arranging

Permutation involves 2 steps- Selection + Arrangement
But combination involves 1 step= only selection

So, I am not able to understand why we need to divide by 3!.

Please assist where is my reasoning wrong.


When you "select" using the combinations formula multiple times on the same group, you are arranging by default.

There are 8 cars.
You are selecting one using 8C1. Say you select Red model A.
Next using 6C1 you select Blue model A.
Next, using 4C1 you select Green model B.
This is one way of selection.


In another case,
Using 8C1, say you select Blue model A.
Next using 6C1, you select Red model A.
Next, using 4C1 you select Green model B.
This is another way of selection.

These will be counted as 2 selections even though the cars selected are the same: Blue model A, Red model A and Green model B.

So when the selection is made by selecting multiple times from the same bunch, you do end up arranging. Hence you need to divide by 3!.

As for 8P1, note that this is the number of ways to select 1 car out of 8 and then arrange the selected 1 car. It doesn't make any sense because you cannot "arrange" one object.
Actually 8C1 and 8P1 are the same: 8!/7!

Check out my last week's post. You might find it useful.
http://www.veritasprep.com/blog/2016/01 ... mbination/
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New post 26 Mar 2012, 19:31
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Bunuel, I don't understand the logic behind why you grouped it like that.
in MGMAT they said to use the slot method, first car has 8 choices, second has 6, third has 6. to account for overcounting we divide by the factorial 3!. same number but i understand that logic a little better but i am inclined to use your method on the test.
Can you explain?
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New post 09 May 2012, 07:29
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ashish8 wrote:
calreg11 wrote:
Bunuel, I don't understand the logic behind why you grouped it like that.
in MGMAT they said to use the slot method, first car has 8 choices, second has 6, third has 6. to account for overcounting we divide by the factorial 3!. same number but i understand that logic a little better but i am inclined to use your method on the test.
Can you explain?


Bunuel, Calgreg, if I follow the MGMAT method I don't understand what over counting they are talking about. I understand the 8 for the first choice, 6 for the second, and 4 for the third, but why do the division afterwards?


Check this problem: if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html and this post there: if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html#p775925
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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 27 Dec 2012, 20:43
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enigma123 wrote:
The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

A. 24
B. 32
C. 48
D. 60
E. 192

Any idea how to solve this guys? I don't have an OA unfortunately.


How many ways to select either Model A or Model B for three cars? 2*2*2 = 8
How many ways to select distinct colors for 3 cars? 4!/3!1! (NOTE: Order doesn't matter) = 4

4*8 = 32

Answer: B
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New post 01 Aug 2014, 03:06
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enigma123 wrote:
The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

A. 24
B. 32
C. 48
D. 60
E. 192


I approached this question as follows:

1) Identified the total number of possible selections (including with two same colors).
We need to select 3 cars out of 8.
8! / 3!*5! = 56 options

2) Deducted the number of options when two cars of the same color are selected:
If we select, for example, green & green => 6 cars for the third option are left => so we need to deduct this 6 options.
We have 4 different colors => 6 * 4 = 24
56 - 24 = 32 (D)
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New post 08 Nov 2014, 09:27
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Avisek47 wrote:
The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

(A) 24
(B) 32
(C) 48
(D) 60
(E) 192


Case 1) all the three selected cars are from model A =4C3=4
Case 2) all the three selected cars are from model B= 4C3=4
Case 3) 1 car from model A and 2 car from model B = 4C1*3C2(because all three cars must have different colors)=12
Case 4) 1 car from model B and 2 car from model A = 4C1*3C2=12

thus total number of combinations= 4+4+12+12=32
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New post 23 Aug 2015, 10:53
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I found the slot method to be simple to use for this question. There are 8 options for Car 1, 6 options for Car 2, and 4 options for Car 3. For example, if I selected A_Red for Car 1, I cannot select B_Red for Car 2 and so forth. Finally, the order in which I select the cars does not matter so I divide the number by 3!.

The answer then becomes: 8*6*4/3! = 32 (B)
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New post 29 May 2019, 23:32
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rishabhjain13 wrote:
Hi Karishma,

Why do we need to 'unarrange' the cars, since with 8C1.6C1.4C1, we have only picked three cars and the 1st car picked can be put at any slot (1/2/3), but since we are not concerned with the arrangement, we have anyway not multiplied the expression by 3!, and hence, left it unarranged.

I'm sorry, I know I am getting confused here, but would be thankful if you could clarify this a bit further.

Thank you.


When we say 8C1*6C1*4C1, we have already arranged them in slot1, slot2 and slot 3 without even multiplying by 3!.

Here is why:

Say these are your 8 cars: BlackA, BlackB, GreenA, GreenB, RedA, RedB, BlueA, BlueB

Now, you do 8C1 and select BlackA.

Leftover cars: GreenA, GreenB, RedA, RedB, BlueA, BlueB

Of these 6, you do 6C1 and select GreenB

Leftover cars: RedA, RedB, BlueA, BlueB

Of these 4, you do 4C1 and select RedA

- Done

Now consider another case:

8 cars: BlackA, BlackB, GreenA, GreenB, RedA, RedB, BlueA, BlueB

Now, you do 8C1 and select RedA.

Leftover cars: BlackA, BlackB, GreenA, GreenB, BlueA, BlueB

Of these 6, you do 6C1 and select GreenB

Leftover cars: BlackA, BlackB, BlueA, BlueB

Of these 4, you do 4C1 and select BlackA

- Done

The two cases are different but what you got from them is the same {BlackA, RedA, GreenB}. You cannot count them twice.

The method of 8C1 * 6C1 * 4C1 is the basic counting principle in which you are automatically putting them in slot1, slot2 and slot3. For only selection, you do need to un-arrange by dividing by 3!.

Check out these posts too:
https://www.veritasprep.com/blog/2011/1 ... inatorics/
https://www.veritasprep.com/blog/2011/1 ... binations/
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New post 07 May 2012, 20:46
Bunnel,

I tried solving it through 8C3 = 56, as there are 8 options available Model A with 4 colors, and Model B with 4 colors, and we have to select 3 out of those..Can you please explain what is wrong in this approach ?

Thanks!
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New post 07 May 2012, 23:44
gmihir wrote:
Bunnel,

I tried solving it through 8C3 = 56, as there are 8 options available Model A with 4 colors, and Model B with 4 colors, and we have to select 3 out of those..Can you please explain what is wrong in this approach ?

Thanks!


8C3 gives total # of ways to select 3 cars out of 8. But in the question we have a restriction saying that "all the cars (selected) are to be different colors", naturally restriction will reduce the number of combinations possible, so 56 cannot be a correct answer.

Hope it's clear.
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New post 09 May 2012, 07:21
calreg11 wrote:
Bunuel, I don't understand the logic behind why you grouped it like that.
in MGMAT they said to use the slot method, first car has 8 choices, second has 6, third has 6. to account for overcounting we divide by the factorial 3!. same number but i understand that logic a little better but i am inclined to use your method on the test.
Can you explain?


Bunuel, Calgreg, if I follow the MGMAT method I don't understand what over counting they are talking about. I understand the 8 for the first choice, 6 for the second, and 4 for the third, but why do the division afterwards?
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