3 cars selection

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karthikpandian19: Legendary Member; Posts: 1665; Joined: Thu Nov 03, 2011 7:04 pmFollowed by: 70 members; Thanked: 165 times

by karthikpandian19 » Mon Jan 09, 2012 6:06 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

If Explanation is provided with the answer it would be great

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Source: Beat The GMAT — Problem Solving | Mon Jan 09, 2012 6:06 pm

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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 » Mon Jan 09, 2012 7:20 pm

karthikpandian19 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

If Explanation is provided with the answer it would be great

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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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Jan 10, 2012 4:54 am

I posted solutions here:

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LeoBen: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Mon Oct 31, 2011 4:41 am; Thanked: 2 times

by LeoBen » Tue Jan 10, 2012 5:37 am

Slot method

Car 1 X 2 X 3 = _ X _ X _

so 8 X 6 X 4.

The above can be arranged in 3! ways, hence reduces the no. of ways --> 8X6X4/3! = 32

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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 » Tue Jan 10, 2012 7:48 am

karthikpandian19 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

If Explanation is provided with the answer it would be great

Here's another possible approach (using the "slot" method):

First select 3 separate 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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satishchandra: Master | Next Rank: 500 Posts; Posts: 172; Joined: Fri Oct 21, 2011 9:10 pmFollowed by: 2 members; Thanked: 7 times

by satishchandra » Wed Jan 11, 2012 12:21 am

karthikpandian19 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

If Explanation is provided with the answer it would be great

8C3-4*6 = 32 ways

Total number of ways to pick 3 cars out of 8 choices = 8C3
Number ways to select two cars of same color in three car selection = 24

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shubhamkumar: Senior | Next Rank: 100 Posts; Posts: 72; Joined: Mon Apr 11, 2011 6:45 pm; Thanked: 2 times

by shubhamkumar » Mon Apr 09, 2012 10:23 am

satishchandra wrote:
karthikpandian19 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

If Explanation is provided with the answer it would be great
8C3-4*6 = 32 ways

Total number of ways to pick 3 cars out of 8 choices = 8C3
Number ways to select two cars of same color in three car selection = 24

Lifted the following from a blog on MGMAT.
https://www.manhattangmat.com/forums/the ... 12659.html
alternative approach:

no. of groups not having any two cars of same colour=total no. of possible groups of 3 cars -no.of groups of 3 cars having any two cars of same colour

no.of groups of 3 cars having any two cars of same colour
=6 . 4= 24
scenario1: black_ black_ (any one colour from rest of 6 cars)
= 6 possible ways(order does not matter in group selection)

similarly,all total 4 scenarios for black ,blue ,green ,and red colour respectively,leading to 6.4 i.e 24 ways of selection.

no. of groups not having any two cars of same colour=total no. of possible groups of 3 cars -no.of groups of 3 cars having any two cars of same colour
=8c3 -24
=56-24
=32
hope it helps

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shubhamkumar: Senior | Next Rank: 100 Posts; Posts: 72; Joined: Mon Apr 11, 2011 6:45 pm; Thanked: 2 times

by shubhamkumar » Mon Apr 09, 2012 11:02 am

Anurag@Gurome wrote:
karthikpandian19 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

If Explanation is provided with the answer it would be great
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.

I had a hell of a problem understanding why the combination had to be divided by 3!.
I went back to the basics and found something which helped me understand.Putting it down for people who may have been in the same boat.
Well we know that the only difference between permutation and combination:
Permutation :arrow: is when order does not matter=nPr
Combination :arrow: is when order does matter =nCr
If we write down the mathematical notations of both we know that
nPr=n!/(n-r)! :arrow: order does not matter i.e. all the possible combinations.But when we want to find the total number of combinations :arrow: order does matter,we divide the Number of permutations by r! or nCr=nPr/n!=n!/r!(n-r)!
Therefore in the problem:

Hence, # of possible combinations = (8 * 6 * 4)/3! = 32

8*6*4 actually is the number of permutations.And then converting it into a combination by dividing by r!,which in this case is 3!.
Phew!
For those who want to find how nCr=nPr/r! is derived,read the following article.
https://www.mathsisfun.com/combinatorics ... tions.html

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