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Difference between revisions of "2004 AMC 10A Problems/Problem 12"

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==Problem==
 
==Problem==
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Henry's Hamburger Heaven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions.  A customer can choose one, two, or three meat patties, and any collection of condiments.  How many different kinds of hamburgers can be ordered?
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Henry's Hamburger Haven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions.  A customer can choose one, two,or three meat patties and any collection of condiments.  How many different kinds of hamburgers can be ordered?
  
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<math> \mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 256 \qquad \mathrm{(C) \ } 768 \qquad \mathrm{(D) \ } 40,320 \qquad \mathrm{(E) \ } 120,960  </math>
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<math> \text{(A) \ } 24 \qquad \text{(B) \ } 256 \qquad \text{(C) \ } 768 \qquad \text{(D) \ } 40,320 \qquad \text{(E) \ } 120,960  </math>
  
 
==Solution==
 
==Solution==
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For each condiment, our customer may either order it or not.  There are 8 condiments.  Therefore, there are <math>2^8=256</math> ways to order the condiments.
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Think of the condiments as in a set with 8 elements.  There are <math>8</math> total condiments to choose from.  Therefore, there are <math>2^8=256</math> ways to order the condiments. There are also <math>3</math> choices for the meat, making a total of <math>256\times3=768</math> possible hamburgers. <math>\boxed{\mathrm{(C)}\ 768}</math>
  
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There are also 3 choices for the meat, making a total of <math>256\times3=768</math> possible hamburgers <math>\Rightarrow\mathrm{(C)}</math>.
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== Video Solution by OmegaLearn ==
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https://youtu.be/0W3VmFp55cM?t=373
  
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==See Also==
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~ pi_is_3.14
  
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*[[2004 AMC 10A Problems]]
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== Video Solution ==
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https://youtu.be/3MiGotKnC_U?t=950
  
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*[[2004 AMC 10A Problems/Problem 11|Previous Problem]]
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~ ThePuzzlr
  
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*[[2004 AMC 10A Problems/Problem 13|Next Problem]]
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==Video Solution==
 +
https://youtu.be/EIExyf8U7O0
 +
 
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Education, the Study of Everything
 +
 
 +
==Video Solution==
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https://youtu.be/j-jNtSwTrxA?t=241 - AMBRIGGS
 +
 
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== See also ==
 +
{{AMC10 box|year=2004|ab=A|num-b=11|num-a=13}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:03, 8 August 2024

Problem

Henry's Hamburger Haven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two,or three meat patties and any collection of condiments. How many different kinds of hamburgers can be ordered?

$\text{(A) \ } 24 \qquad \text{(B) \ } 256 \qquad \text{(C) \ } 768 \qquad \text{(D) \ } 40,320 \qquad \text{(E) \ } 120,960$

Solution

Think of the condiments as in a set with 8 elements. There are $8$ total condiments to choose from. Therefore, there are $2^8=256$ ways to order the condiments. There are also $3$ choices for the meat, making a total of $256\times3=768$ possible hamburgers. $\boxed{\mathrm{(C)}\ 768}$

Video Solution by OmegaLearn

https://youtu.be/0W3VmFp55cM?t=373

~ pi_is_3.14

Video Solution

https://youtu.be/3MiGotKnC_U?t=950

~ ThePuzzlr

Video Solution

https://youtu.be/EIExyf8U7O0

Education, the Study of Everything

Video Solution

https://youtu.be/j-jNtSwTrxA?t=241 - AMBRIGGS

See also

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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