The Carson family will purchase three used cars. There are

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.

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

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.

RELATED VIDEOS

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

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

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

Yes, I think you need to go through a PnC book from scratch.

"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.

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

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.
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2016/01 ... mbination/

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?

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.

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/

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.

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!

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?

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?

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