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rupsk: Master | Next Rank: 500 Posts; Posts: 123; Joined: Mon Feb 07, 2011 12:11 pmFollowed by: 1 member

by rupsk » Thu Dec 22, 2011 3:53 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?

24

32

48

60

192

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Source: Beat The GMAT — Problem Solving | Thu Dec 22, 2011 3:53 pm

neelgandham: Community Manager; Posts: 1060; Joined: Fri May 13, 2011 6:46 amFollowed by: 52 members; Location: Utrecht, The Netherlands; Thanked: 318 times

by neelgandham » Thu Dec 22, 2011 4:07 pm

Total types of cars

Model A - Blue
Model A - Black
Model A - Red
Model A - Green
Model B - Blue
Model B - Black
Model B - Red
Model B - Green

Number of ways of selecting the first car : 8(let us assume that the family selected a Red Car, then the family cannot select another Red car. So, Only 6 cars are left for the next seleionction)

Number of ways of selecting the Second car : 6(let us assume that the family selected a Blue Car, then the family cannot select another Blue car. So, Only 4 cars are left for the next selection)

Number of ways of selecting the Third car : 4

Total number of ways = c

But a car combination
Red-Green-Blue is same as
Red-Blue-Green, which is same as
Green-Blue-Red, which is same as
Green-Red-Blue, which is same as
Blue-Red-Green, which is same as
Blue-Green-Red.

So, same combination repeats 6 times, so the number combinations of three cars can the Carsons select if all the cars are to be different colors? = 8*6*4/6 = 32

Anil Gandham
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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Dec 22, 2011 4:16 pm

rupsk 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?

24

32

48

60

192

First select 3 colors. This can be accomplished in 4C3 ways (4 ways).
For the first color, choose a model (A or B) this can be accomplished in 2 ways.
For the second color, choose a model (A or B) this can be accomplished in 2 ways.
For the third color, choose a model (A or B) this can be accomplished in 2 ways.

The total number of ways to accomplish all four of these steps = 4x2x2x2=32 = B

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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rupsk: Master | Next Rank: 500 Posts; Posts: 123; Joined: Mon Feb 07, 2011 12:11 pmFollowed by: 1 member

by rupsk » Thu Dec 22, 2011 4:19 pm

thanks

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Thu Dec 22, 2011 6:17 pm

rupsk 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?

24

32

48

60

192

1st car can be selected from 8 cars in 8 ways
2nd car can be selected from 6 cars in 6 ways
3rd car can be selected from 4 cars in 4 ways
Hence, # of possible combinations = (8 * 6 * 4)/3! = 32

The correct answer is B.

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ArunangsuSahu: Master | Next Rank: 500 Posts; Posts: 382; Joined: Thu Mar 31, 2011 5:47 pm; Thanked: 15 times

by ArunangsuSahu » Fri Dec 30, 2011 11:05 am

Here Model is fewer than the number of Cars to be purchased.

Let's break down the problem

[color=blue]CASE I:[/color]

All 3 colors are of one model

So Combination is 4C3= 4 and there are 2 models so 4*2 =8

CASE II:

One cars is of color of one model and the remaining two are of different colors of the 2nd model = 4C1*3C2=12. This also will happen for 2 models. So total 2*12=24

Combining cas I and II . Total Combinations = 8+24 = 32

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