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The Carson family will purchase three used cars. There are

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wtstone
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Re: The Carson family will purchase three used cars. There are [#permalink] The Carson family will purchase three used cars. There are   12 Sep 2022, 13:50
Bunuel
enigma123
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

Any idea how to solve this guys? I don't have an OA unfortunately.

\(C^3_4*2^3=4*8=32\), (\(C^3_4\) selecting 3 different colors from 4 and multiplying by 2*2*2=2^3 since each color car has two options: model A or model B).

Answer: B.

Similar question to practice:
https://gmatclub.com/forum/if-there-are- ... 99992.html
https://gmatclub.com/forum/if-a-committe ... 98533.html

Hope it helps.

I need some help with this one, please. I have come up with 16 as the answer given the following:

with there being 2 different types of cars, and the family needing a combination of 3, there are a total of 4 possibilities:

AAA
BBB
AAB
ABB

if this were a permutation problem then I could understand how you arrive at 8 permutations of cars, but I only get the 4.

can anyone explain?
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Re: The Carson family will purchase three used cars. There are [#permalink] The Carson family will purchase three used cars. There are   12 Sep 2022, 13:53
Bunuel
enigma123
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

Any idea how to solve this guys? I don't have an OA unfortunately.

\(C^3_4*2^3=4*8=32\), (\(C^3_4\) selecting 3 different colors from 4 and multiplying by 2*2*2=2^3 since each color car has two options: model A or model B).

Answer: B.

Similar question to practice:
https://gmatclub.com/forum/if-there-are- ... 99992.html
https://gmatclub.com/forum/if-a-committe ... 98533.html

Hope it helps.

I need some help with this one, please. I have come up with 16 as the answer given the following:

with there being 2 different types of cars, and the family needing a combination of 3, there are a total of 4 possibilities:

AAA
BBB
AAB
ABB

if this were a permutation problem then I would arrive at 8 possibilities. Can anyone explian?
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Re: The Carson family will purchase three used cars. There are [#permalink] The Carson family will purchase three used cars. There are   02 Oct 2022, 05:07
My approach:

We have total 8 cars, model A (blue, black, red, and green) & model B (blue, black, red, and green).
We can select cars in two ways:

1) 3 cars from model A (blue, black, red, and green) & 0 cars from model B (blue, black, red, and green) or 0 cars from model A (blue, black, red, and green) & 3 cars from model B (blue, black, red, and green).
2) 1 car from model A (blue, black, red, and green) & 2 cars from model B (blue, black, red, and green) or 2 car from model A (blue, black, red, and green) & 1 cars from model B (blue, black, red, and green)

Now selection
For 1) : 4C3*4C0 + 4C3*4C0 = 2* 4C3 = 8
For 2) : 4C1*3C2 + 4C1*3C2 = 2* 4C1*3C2 = 24 (3C2 coz we are only left with three options after choosing one car from other group)

Hence answer = 8+24 = 32
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Re: The Carson family will purchase three used cars. There are [#permalink] The Carson family will purchase three used cars. There are   29 Nov 2022, 23:33
KarishmaB
ashish8
calreg11
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!


Hi karishma
As per my understanding we use permutation "P" for arrangements and use combination "C" where the order doesn't matter.
For this question
if I have to choose cars without considering order, I can choose it use Combination
so
total = 8C1 x 6C1 x 4C1
Now since I have used combination instead of Permutation, why do I have to divide by 3! to compensate for order?

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Re: The Carson family will purchase three used cars. There are [#permalink] The Carson family will purchase three used cars. There are   24 May 2023, 06:27
We need to choose a COMBINATION of 3 cars (each of different color) from a selection of 8 cars.

Using the slot method:
8 choices for color #1 x 6 choices for color #2 x 4 choices for color #3 = 192

The slot method will produced an ordered result - it has overcounted by 3! (I call this factor "slots factorial" as it will always mirror the number of slots), so we can divide by the same factorial to get rid of the duplicate counting.

192/3! = 32
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The Carson family will purchase three used cars. There are [#permalink] The Carson family will purchase three used cars. There are   31 Aug 2023, 05:37
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enigma123
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

select 3 colors out of 4 in 4C3 ways
Select one model out of two in each color in 2C1*2C1*2C1 ways

Final Outcome = 4C3*2C1*2C1*2C1 = 32

Answer: Option B
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Re: The Carson family will purchase three used cars. There are [#permalink] The Carson family will purchase three used cars. There are   26 Sep 2024, 02:31
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Re: The Carson family will purchase three used cars. There are [#permalink]
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