Carson family from Manhattan

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AbhishekVerma416: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Tue Jun 07, 2011 12:20 pm

by AbhishekVerma416 » Thu Jun 16, 2011 4:22 pm

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

..

OA: B
The answer given is B. I would like to understand why its B and not E.

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Source: Beat The GMAT — Problem Solving | Thu Jun 16, 2011 4:22 pm

winniethepooh: Master | Next Rank: 500 Posts; Posts: 370; Joined: Sat Jun 11, 2011 8:50 pmFollowed by: 2 members; Location: Arlington, MA.; Thanked: 27 times

by winniethepooh » Thu Jun 16, 2011 5:47 pm

I don't know how you get the answer to be 192.

According to me you have 4C3 to select the color of the cars = 4!/3! = 4ways

Now, out of these 4 combinations you have to choose 3 cars of different colors.Each color can be selected in 2 ways(as there are 2 models). So, you have 2x2x2 = 2^3 ways of getting 3 colors.

Total numbers of combinations of three cars of different colors = 4 x 2^3
= 32 ways.

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amit2k9: Master | Next Rank: 500 Posts; Posts: 461; Joined: Tue May 10, 2011 9:09 amFollowed by: 3 members; Location: pune; Thanked: 36 times

by amit2k9 » Thu Jun 16, 2011 7:32 pm

total cars = 8
3 cars of different colors = 8*6*4 = 192

since the order is not particular here = 3!
192/3! = 32

A similar type - https://www.manhattangmat.com/forums/a-c ... t9945.html

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