The Carson family will purchase three used cars. There are

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}

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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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.

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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/1 ... inatorics/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/1 ... binations/

If they select only one car model, then they have 4C3 = 4 ways to choose Model A cars and another 4 ways to choose Model B cars, for a total of 8 ways.

If they select both models of cars, they have 4C2 = 6 ways to choose 2 Model A cars and 2C1 = 2 ways to choose 1 Model B car, for a total of 6 x 2 = 12 ways to choose 2 Model A and 1 Model B cars. Likewise, they have a total of 12 ways to choose 2 Model B and 1 Model A cars. Therefore, they have a total of 24 ways to choose both models of cars.

Therefore, they have 8 + 24 = 32 ways to purchase 3 cars with all different colors.

Alternate Solution:

For the first car, there are 4 x 2 = 8 choices. Since the second car cannot be the same color as the first, there are 8 - 2 = 6 choices for the second car. Similarly, since the last car cannot be of the same color as the first two, there are 8 - 4 = 4 choices for the last car. Since the order in which the cars are being chosen does not matter, the number of ways to make the purchase is (8 * 6 * 4)/3! = 8 * 4 = 32.

Answer: B

Here's what I did if it helps anyone understand

4 Model A cars
4 Model B cars
Need to pick 3

All from A: 4C3 = 4
All from B: 4C3 = 4
2 of A and 1 of B: 4C2 * 2 (we can't overlap colors so the 2 colors we picked from A are out from B leaving 2 options only) = 12
2 of B and 1 of A: 4C2 * 2 (same logic as above) = 12

= 4+4+12+12 = 32

The question asks "How many different combinations of three cars can the Carsons select if all the cars are to be different colors?" so it does not specfify that the cars have to be differerent Models.

I get the #arrangements for 3 of 4 colors is equal to 4, I get only 4 options to arrange 3 cars if they can be THE SAME model:

AAA
BBB
ABA
ABB

since the order is not specified, it should be only 4 and not 8 ways, which would give 4 x 4 = 16 arrangements and not 32?

Interesting, I just write it down again and I get it ))
I also thought it’s 4 models combination, but :

AAA
ABA
AAB
ABB

BBB
BAB
BBA
BAA

Posted from my mobile device

They can choose either of (1 Model A + 2 Model B) OR (2 Model A + 1 Model B) OR (3 Model A) OR (3 Model B)

(1 Model A + 2 Model B) OR (2 Model A + 1 Model B) = 2 * 4C1 * 3C2

(3 Model A) OR (3 Model B) = 2 * 4C3

12 + 12 + 4 + 4 = 32

I approached this question by the following way:

C21*3=8

ABA,BAB,AAB,BBA,BAA,ABB,BBB,AAA

For example: ABA
1)When select A: C41 (four colors available)
2)B:C31
3)A:C21

so we have 24 ways to select colors.

I know the answer is 32.

I hope someone can tell me what I do wrong.

You're double counting in two different ways.

ABA is the same as AAB because the question is asking for groups of cars and is not interested in the different ways a given group can be created.

Eliminating those duplicates leaves just 4 groups of cars:

One with 2As and 1B.
One with 2Bs and 1A
One with 3As
One with 3Bs

Notice I didn't write out the groups because I don't want to imply that the order in which I wrote them makes any difference.

The second instance of double counting is in assigning colors.

Yes, there are 4 ways of selecting 3 colors to be assigned.

But in assigning colors the way that you are for the two groups with both A's and B's, your method counts a Red A car and a Green A car, for example, twice.

The easiest way to assign colors for the groups with two of the same model is to assign a color to the different model, 3 ways, and then the duplicate model's colors are taken care of, 1 way.

So there are 3 color schemes for both the two A and two B groups * 4 color choices each equals

24

The 3A and 3B group each have 4 color choices for a total of

8

So the overall total is 32


Now, you can challenge my statement about double counting ABA and AAB, for example, because if you DO follow that approach, you can view the 3 different orders of those letters as assigning the colors 3 ways as I describe above and you will get the correct answer.

However, that approach combines into one step the selection of cars and assigning of colors which I think can be confusing.

Posted from my mobile device

Regor60
Hi, Regor60!

Thank you for your kind explanation!

I had hard time to understand the explanation to be honest. You said that I doubled counted the ways. But I was thinking that an another way to get the answer: 2^3*C43. In this way, I still count 8 ways, but I assign colors by choosing 3 out of 4. I am not sure if this approach is acceptable.

I get confused because I don't know when I should use the combination formula.

Yes that approach is fine and is what I refer to in the second part of my post

Posted from my mobile device

FWIW, the models are distractors here. There are no limitations on the model, only on the color. Effectively, we just have eight cars: red, red, black, black, blue, blue, green, green. We are prohibited from picking both cars of any color.

We have 8 cars from which to choose.
There are 8 options for the first selection.
Once we make the first selection, we can no longer pick that same model/color combination and we can't pick the other model of the same color, so there are 6 options.
Once we make the first two selections, we can no longer pick either of those exact model/color combinations and we can't pick the other models of the same colors, so there are 4 options.
8*6*4 = 192
But picking Ared, Bblack, Bgreen is the same as picking Bblack, Bgreen, Ared, so we need to eliminate duplicates. There are 3 options for car was chosen first, 2 for second, and 1 for third.
3*2*1 = 6

192/6 = 32

Answer choice B.

A simple method is to visual the ways of making the 3 purchases.
The Carson family could:
a. buy all 3 cars of the same model
b. buy 2 of one model and 1 of another model

In a, we have 4C3 ways of choosing 3 different colours of a model. Multiply it by 2 to account for the two models available -> 4C3*2 = 8
In b, we have 4C2 ways of choosing 2 different colours of a model and 2C1 ways of choosing the third colour from the other model (as we remove the two colours that are already chosen). Multiply it by 2 to account for the two models available -> 4C2*2C1*2 = 24

Add a & b up to get 32 ways.