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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 Sep 2017, 13:16
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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


Thanks a lot for the explanation VeritasPrepKarishma. I was stuck with 24 as solution. Now I realized the conceptual gap.
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New post 29 Sep 2017, 12:48
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


My method in solving this was a little different.

First, there are 8 cars to choose from and we need to pick 3 cars. This can be done in 8!/5!*3!=56 ways.
Next, let's choose all cars of the same color. We have 4 colors and 2 models. There is no way all 3 cars can be of the same color.
Next, let's choose 2 cars with the same color. let's assume we have chosen both the red cars, now we need only 1 car. the way we choose this 1 car is 6!/5!*1!=6.
but we could have just as easily chosen any of the other 3 colors for our first 2 cars, so we multiply 6*4=24.

now, since the problem asks us to find the combinations of cars that have a different color each, we need to subtract the number of cars with the same color from the total combinations. i.e. 56-24=32.

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

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New post 21 Apr 2018, 07:21
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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


Take the task of selecting cars and break it into stages.

Stage 1: Select 3 different colors.
Since the order in which we select the colors does not matter, we can use combinations.
We can select 3 colors from 4 colors in 4C3 ways (4 ways).

ASIDE: If anyone is interested, we have a video on calculating combinations (like 4C3) in your head (see below)

Stage 2: For one color, choose a model
There are two models (A or B) so this stage can be accomplished in 2 ways.

Stage 3: For another color, choose a model
There are two models (A or B) so this stage can be accomplished in 2 ways.

Stage 4: For the last remaining color, choose a model
There are two models (A or B) so this stage can be accomplished in 2 ways.

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus select the 3 cars) in (4)(2)(2)(2) ways (= 32 ways)

Answer: B

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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

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New post 21 May 2018, 06:15
VeritasPrepKarishma
Please tell where am i going wrong in my concepts?

We need to select 3 colors out of 4 colors.There are two model of cars too!

So, [C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)]+[C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)]
-------Model A first,Model B 2nd-------- -------Model B first,Model A 2nd--------

Thus, on calculating we get: 28+28=56
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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 21 May 2018, 22:50
We need to select 3 Cars from 4 available = 3C4 = 4

Select 1 car from each color = 8

Total combination = 4*8 = 32

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

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New post 21 May 2018, 23:07
raunakme19 wrote:
VeritasPrepKarishma
Please tell where am i going wrong in my concepts?

We need to select 3 colors out of 4 colors.There are two model of cars too!

So, [C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)]+[C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)]
-------Model A first,Model B 2nd-------- -------Model B first,Model A 2nd--------

Thus, on calculating we get: 28+28=56


I am not sure of the logic you have followed here.
Using 4C3, you select 3 colours of the 4. Then why do you use 4C2 and 4C1?
Also each colour is available in 2 different models so next set is to pick a model for each colour.
Say you picked blue, black and red.
For blue, you pick either model A or B so 2 options. For black you pick either model A or B so 2 options. For red, you pick either model A or B so 2 options.

In all, 4C3 * 2 * 2 *2 ways.

Also, it is just a selection. No arrangement involved so first second order is immaterial.
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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 22 May 2018, 12:25
VeritasPrepKarishma wrote:
raunakme19 wrote:
VeritasPrepKarishma
Please tell where am i going wrong in my concepts?

We need to select 3 colors out of 4 colors.There are two model of cars too!

So, [C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)]+[C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)]
-------Model A first,Model B 2nd-------- -------Model B first,Model A 2nd--------

Thus, on calculating we get: 28+28=56


I am not sure of the logic you have followed here.
Using 4C3, you select 3 colours of the 4. Then why do you use 4C2 and 4C1?
Also each colour is available in 2 different models so next set is to pick a model for each colour.
Say you picked blue, black and red.
For blue, you pick either model A or B so 2 options. For black you pick either model A or B so 2 options. For red, you pick either model A or B so 2 options.

In all, 4C3 * 2 * 2 *2 ways.

Also, it is just a selection. No arrangement involved so first second order is immaterial.





Thanks Karishma!!
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New post 25 Aug 2018, 16:13
I understand that 1C8* 1C6 *1C4 = 8*6*4 is "trying to arrange the sequence btw 3 cars" so we need to "un-arrange them" by dividing 8*6*4 over 3!/1!2! (which equals to 3), so the whole equation is "8*6*4 then divide by 3"= 64.
BUT,
the official correct answer is 32.

Could someone explain?

Thx a lot!!!
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New post 01 Sep 2018, 05:50
Its straightforward:-

Either all three of one model which means 4C3 ways for model A and 4C3 for model B; OR

2 could be selected from model A and one from model B and vice versa but then the precaution which one needs to take is that all must be of different colours.

Therefore it will be 4C2*2C1 for 2 of Model A AND one of model B OR 4C2*2C1 for 2 of Model B AND one of model A.

Therefore total ways is: 2*4C3 + 2*4C2*2C1 = 2*4 + 2*6*2 = 32 ways. (ANS)
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New post 01 Sep 2018, 06:21
We have 8 types of cars:Model A (B, B, R & G) + Model B (B,B, R & G).
Let's select 3 cars of different colors from these 8 choices.
1st selection - 8 choices
2nd selection- 6 choices (As one color selection already done above, let's say Model A - Black. This put restrictions on model B - Black as well as we need different colors).
3rd selection - 4 choices (As 2 colors already selected above so we are left with only 2 colors of models A&B).

Using FCP, Number of combinations is 8X6X4/3! = 32. Choice B.
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New post 11 Mar 2019, 06:38
Hi,

As some fellas here, I wasn't able to understand Bunuel's logic (Sorry Bunuel, I wish I could be like you).

Therefore, here is my own one, it leads to the correct answer as well.

Number of ways of selecting 3 cars over 8: \(\frac{8!}{3!5!}=56\)

Number of ways of selecting at least two blue cars: First car blue and Second car blue \(*3\) (rest of the colors available) \(*2\) (number of models)\(=6\)

Thus, number of ways of selecting at least two cars of the same color \(6*4=24\) (where 4 is the number of colors)

And, finally, number of ways of selecting 3 cars of different colors \(56-24=32\)

Answer B


Hope it helps!
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New post 28 May 2019, 20:37
VeritasKarishma 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 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.
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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 30 May 2019, 00:32
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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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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 27 Jun 2019, 02:22
3 diff color cars from Type A= 4C3 =4
3 diff color cars from Type B= 4C3 =4
2 cars from Type A and 1 car from Type B= 4C2 * 2C1(1 color selected from 2 left over Type B colors) = 6*2=12
1 cars from Type A and 2 car from Type B= 4C1 * 3C2(2 color selected from 3 left over Type B colors) = 4*3=12
Total= 32
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Re: The Carson family will purchase three used cars. There are  [#permalink]

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New post 10 Jul 2019, 23:32
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


Out of 4 colour options, they can choose 3 different colours in 4C3 = 4 ways.
Now each colour has two different models, so models can be selected in 2X2X2 = 8 ways.
Total ways to choose 3 cars = 4X8 = 32 ways
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Re: The Carson family will purchase three used cars. There are   [#permalink] 10 Jul 2019, 23:32

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