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The Carson family will purchase three used cars. There are

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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, 00:51
VeritasPrepKarishma wrote:
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


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
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Re: The Carson family will purchase three used cars.  [#permalink]

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New post 18 Oct 2013, 09:41
I get an answer of 12. 3C2 (3 cars out of 2)=3, 4C3 (4 colors, 3 cars)=4, 4*3=12. I think the provided answers are all wrong, or maybe I' missign something. I've been drilling this type of program for 6 weeks, and either I haven't learned a thing, or the answers are incorrect.
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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 17 May 2014, 13:43
VeritasPrepKarishma wrote:
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.


Hi Karishma,

I made a similar error in the beginning and looking back at it, it might have been because I might have tried to mix in "probability" into the combinatorics problem.

If you treat it as a probability, wouldn't the above posters approach be correct?
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New post 18 May 2014, 21:33
russ9 wrote:
VeritasPrepKarishma wrote:
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.


Hi Karishma,

I made a similar error in the beginning and looking back at it, it might have been because I might have tried to mix in "probability" into the combinatorics problem.

If you treat it as a probability, wouldn't the above posters approach be correct?


The approach of the poster here is quite wrong. You just need the number of cases here so it is not a probability question. If you do want to figure out the correct method in case it were a probability question, give the question you want to discuss and the solution you will use. Then I can tell you whether you are correct.
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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 27 Jun 2014, 12:51
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
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Re: The Carson family will purchase three used cars. There are two models  [#permalink]

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New post 08 Nov 2014, 15:42
Bunuel can you explain why we can't use the following method:

I can choose any 8 cars for my first car
I can choose any 6 cars for my second
I can choose any 4 cars for my third

8x6x4=192
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Re: The Carson family will purchase three used cars. There are two models  [#permalink]

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New post 08 Nov 2014, 21:24
Select 3 colors out of total 4 colors= 4C3= 4
Each color has 2 different models = 2^3=8
Total combinations = 4*8=32
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Re: The Carson family will purchase three used cars. There are two models  [#permalink]

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New post 09 Nov 2014, 00:06
bankerboy30 wrote:
Bunuel can you explain why we can't use the following method:

I can choose any 8 cars for my first car
I can choose any 6 cars for my second
I can choose any 4 cars for my third

8x6x4=192

Assume that you selected Green A Red B and Black A in one selection and Red B Green A and Black A in the second selection. These two selections are same. So sorting (or permutations) work here. You have to divide 192 by factorial of 3. Thus 192/3!=32 ;)
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New post 23 Aug 2015, 11:53
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 26 Oct 2015, 21:45
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.
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New post 02 Nov 2015, 08:00
I understand the slot method to solve this problem. i.e 8*6*4/3! but when I saw this problem first I tried to solve it by subtracting the combinations where colours of the cars are the same. I'm still not able to find the solution through this, can some one point out the mistake?

Total Choices: 8*8*8 = 512(unrestricted, I assume you can pic a same colour same model 3 times)

Choices where two cars are Blue: 2*2*8 = 32 (Say Blue for first two choices, then remaining can be any of the 8 remaining choices)
Since 4 colours = 32*4 = 128


So for all cars to have a different colour = total - two-cars-same - three-cars-same = 512 - 128 = 384

I guess I need to divide in each step by 3! since order doesn't matter which would just mean my answer is 512/3! - 128/3! = 64... Which is still not the answer.. .What am I missing?
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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 04 Nov 2015, 08:44
Hi Bunuel, VeritasPrepKarishma, Engr2012 , manpreetsingh86 , WoundedTiger

I have approached the problem in this manner.

Total cars are 8.

Let's tag these cars Bl(A), Bk(A), R(A), G(A), Bl(B), Bk(B), R(B), G(B)

Now we have 4 families of cars with similar colors: Bl(A), Bl(B) ; Bk(A), Bk(B); R(A), R(B); G(A), G(B)

As we can't chose 2 cars with same colors, we have to pick 3 different colors from available 4 colors.

So, we have 8 cars and we have to select 3 cars with different cars.

The first car - 1 out of 8 => 8c1
Second car - 1 out of 6 => 6c1
Third car - 1 out of 4 => 4c1

This will be 192.

But guys don't stop here.

Imagine we have picked up Bl(A), Bk(B), G(B)....These cars can be selected in any order.

So the no. of ways will be repeated in the above mentioned 192 arrangements.

So we have to divide the replication.

The replication is 3! ways = 6 ways

So finally the number of ways we can select 3 cars out of 8 cars (of different colors) = 192/6 = 32

Short cut:

(8c1 * 6c1 * 4c1)/ 3! = 32

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New post 01 Feb 2016, 11:50
VeritasPrepKarishma wrote:
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/



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.
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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 21 May 2017, 16:39
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


The total number of cars available to pick is 8 (4 colors and 2 sizes). So for the first choice, there are 8 possible options. After the first choice, cars of that color cannot be picked, so the remaining cars available are 6. Same process for the last choice: 4 options.

8*6*4 = 192 choices.

But the order does not matter, so to eliminate the order, I need to divide by the factorial of the number of decisions (3!).

8*6*4/3*2 = 32.
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New post 07 Jul 2017, 19:13
Hi,
We have A model(B,BL,R,G) and B model(B,BL,R,G)
1. selecting all 3 cars from A none from B= 4C3 = 4
2. selecting all 3 cars from B none from A = 4C3 = 4
3. selecting 2 cars from A and 1 car from B = 4C2*2C1(as we should not select the same color as in A model) = 12
4. selecting 1 car from A and 2 cars from B = 4c1*3C2( as we should not select the same color as in A model) = 12
total = 4+4+12+12 = 32

ans : B

Please correct me if I am wrong.
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New post 08 Jul 2017, 05:53
Hi Bunuel

Help me understand why I am wrong
4C1*3C1*2C1*2*2*2
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New post 08 Jul 2017, 06:35
deepudiscover wrote:
Hi Bunuel

Help me understand why I am wrong
4C1*3C1*2C1*2*2*2


I guess with 4C1*3C1*2C1 = 4*3*2 you are selecting 3 colors for cars. This will give duplications because the order of the colors does not matter. {blue, black, red} color selection is the same as {red, blue, black}, so you should divide 4*3*2 by 3! to dis-arrange and you'll get 4, which is the same as 4C3 = 4 in my solution above. You could simply list all possibilities to avoid formula:

{blue, black, red}
{blue, black, green}
{black, red, green}
{blue, red, green}
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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 08 Jul 2017, 11:36
In total 4 colors, each with 2 choices (Car A or B)

Select 1 Car from each color = 2C1 * 2C1 * 2C1 = 8
Select 3 colors from total of 4 = 4C3 = 4

4* 8 = 32

Ans. B
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New post 26 Jul 2017, 23:48
Hi There,
Thought I'd share my method of solving this. This is a bit different from the ones mentioned above.

Total Cars we can pick from = 8
i:e;
Model A:
1.Blue
2.Black
3.Red
4.Green

Model B

5.Blue
6.Black
7.Red
8.Green

The question looks at each car differently, so I have ignored which model they belong to.

STEP 1 - FINDING THE TOTAL NO. OF WAYS TO PICK 3 CARS
Total no. of ways we can pick = 8 x 6 x 4 (ie: 192 ways)

Explanation:

1.To start off with, you have 8 ways you can pick a car.
2.For example one you pick 1.blue, you cant pick 5.blue. This drops the total no. of cars you can pick to 6.
3.Now, once you pick car 2.Red, you cant pick 6. Red. This means you have 4 cars to pick from
4.This leaves you with 8 x 6 x 4 total ways.

STEP 2 - ELIMINATING REPEATING VALUES
1. You have 192 ways of picking 3 cars. However, this questions does not need you to maintain a specific sequence:

Let's assume you picked red, blue, black. The value of 192 also takes into account blue, black, red.
We need to eliminate repeating values

2. No. of repeating values = no. of ways the three cars can be arranged.
Lets take RED, BLUE and BLACK. These three cars can be arranged as -
RED BLUE BLACK
RED BLACK BLUE
BLUE BLACK RED
BLUE RED BLACK
BLACK BLUE RED
BLACK RED BLUE

There are a total of 6 ways in which any three colors can be arranged.

STEP 3 - GETTING THE ANSWER

Ways in which cars can be picked = TOTAL NO. OF WAYS OF SELECTING THE CARS / NO. OF REPEATING VALUES
ie- 192/ 6 = 32.

Hope this helps :)
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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 27 Aug 2017, 14:44
Hello VeritasPrepKarishma, Please reply.

Please could you point out where m i going wrong in my following explanation:

We have 2 cars: A and B.

In order to buy 3 cars we can have AAA, AAB,ABB or BBB. --> 4 ways
Out of 4 colors we can have 3 --> 4*3*2=24

Therefore, total ways=24*4=96

What am i missing here. Please reply.

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