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Can of Worms!

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by Juggernaut_86 » Sun Sep 11, 2011 12:11 pm
A provision store will purchase three cans of worms. There are two models of cans available, Model 1 and Model 2, each of which is available in four colors: yellow, white, gray, and gold. How many different combinations of the three cans can the store select if all the cans are to be different colors?

a) 24

b) 32

c) 48

d) 60

e) 192


Seeking explanation....Thanks!
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Source: — Problem Solving |

by pemdas » Sun Sep 11, 2011 2:23 pm
each model can be four colors, and three cans can be one model but different colors and/or different models and different colors
C(4,3)*3+C(4,3)*C(3,2)=4*3+4*3=24
a
Juggernaut_86 wrote:A provision store will purchase three cans of worms. There are two models of cans available, Model 1 and Model 2, each of which is available in four colors: yellow, white, gray, and gold. How many different combinations of the three cans can the store select if all the cans are to be different colors?

a) 24

b) 32

c) 48

d) 60

e) 192


Seeking explanation....Thanks!
Success doesn't come overnight!
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by Juggernaut_86 » Sun Sep 11, 2011 2:50 pm
Hello Pemdas,

Unfortunately A is not the correct answer.

OA is B.
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by knight247 » Sun Sep 11, 2011 4:05 pm
I would assume that the model selected is irrelevant as the question does not specify whether it is of importance. If that is the case then we have 2 yellow,2 white,2gray and 2 gold. Now we need to select 3 such that each is of a different colour with no regard for their model.

For the first one we can select any one of the 8 so 8 ways
For the second we can select any other color besides the first one. So 6 ways.
For the third one we only have 4 options to select from so 4 ways

8*6*4 Is the total number of permutations. However the order is not important so we need to convert the permutation to a combination. So dividing by 3! we get 32. Hence B
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by jonathan123456 » Sun Sep 11, 2011 5:40 pm
Agree with Knight247
Never say Die!
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by GMATGuruNY » Sun Sep 11, 2011 7:10 pm
Juggernaut_86 wrote:A provision store will purchase three cans of worms. There are two models of cans available, Model 1 and Model 2, each of which is available in four colors: yellow, white, gray, and gold. How many different combinations of the three cans can the store select if all the cans are to be different colors?

a) 24

b) 32

c) 48

d) 60

e) 192


Seeking explanation....Thanks!
2 model options available:
Number of model options for the first can = 2.
Number of model options for the second can = 2.
Number of model options for the third can = 2.
To combine the options for each can, we multiply:
2*2*2 = 8.

4 colors available:
Number of ways to choose 3 colors from 4 choices = 4C3 = 4.

To combine our model options with our color options, we multiply:
8*4 = 32.

The correct answer is B.
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by knight247 » Mon Sep 12, 2011 6:23 am
Hey Mitch,
Can u give me an opinion on the way I've solved it. Is my logic correct or did my answer just coincidentally match up? lol...Hoping you can check it out. Thanks
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by GMATGuruNY » Mon Sep 12, 2011 7:25 am
knight247 wrote:Hey Mitch,
Can u give me an opinion on the way I've solved it. Is my logic correct or did my answer just coincidentally match up? lol...Hoping you can check it out. Thanks
The reasoning in your solution is perfect. Nice work!
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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Student Review #2
Student Review #3
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by leumas » Mon Sep 12, 2011 8:29 am
GMATGuruNY wrote:A provision store will purchase three cans of worms. There are two models of cans available, Model 1 and Model 2, each of which is available in four colors: yellow, white, gray, and gold. How many different combinations of the three cans can the store select if all the cans are to be different colors?

a) 24

b) 32

c) 48

d) 60

e) 192


Seeking explanation....Thanks!

2 model options available:
Number of model options for the first can = 2.
Number of model options for the second can = 2.
Number of model options for the third can = 2.
To combine the options for each can, we multiply:
2*2*2 = 8.

4 colors available:
Number of ways to choose 3 colors from 4 choices = 4C3 = 4.

To combine our model options with our color options, we multiply:
8*4 = 32.

The correct answer is B.
Hi Mitch,

How can we mathematically derive ans for below question?

How many different combinations of the three cans can the store select if all the cans are to be same color?

The common sense says it's 4. but how do we derive mathematically?
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by navami » Mon Sep 12, 2011 10:10 am
Thanks GMATGuruNY
This time no looking back!!!
Navami
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by sl750 » Mon Sep 12, 2011 1:09 pm
GmatGuruNY

How is one supposed to infer that the selection of the model's isn't important in this question?

Thanks
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