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

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01 Aug 2014, 04:06

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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)

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

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08 Nov 2014, 10:27

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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

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

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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

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

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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!.

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

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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.

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.

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.

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

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

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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?

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

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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!

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.
_________________

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!

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

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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!).

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

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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

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:

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

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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.

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

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11 Sep 2017, 13:47

I'm not 100% comfortable in combinatorics yet, but I did it this way (not sure, though, if that's the correct one):

There are 2 models: A and B.

Hence, we have 4 combinations of models: 1) AAA; 2) BBB; 3) AAB; 4) BBA.

1)=2)= 4C3*2 = 8. As in, we must choose 3 colours out of 4 for the same model, then we multiply by 2, since we have models A and B.

3)=4) = 4C2*2C1*2 = 24. As in, we must choose 2 colours out of 4 for the model that was bought 2 times, then 1 colour out of the remaining 2 for the other model. Since we have 2 ways of grouping the 2 models, we multiply by 2.