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2004 AMC 10A Problems/Problem 12: Difference between revisions

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


<math> \mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 256 \qquad \mathrm{(C) \ } 768 \qquad \mathrm{(D) \ } 40,320 {2}\qquad \mathrm{(E) \ } 120,960  </math>
<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==
A costomer can either order a condiment, or not, and there are 8 condiments.  Therefore, there are <math>2^8=256</math> ways to order the condiments.
Think of the condiments as 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. (You have two choices for each condiment- one choice is to include that condiment, and the other choice is to not include that condiment.) 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>


There are also 3 choices for the meat, making a total of <math>256\times3=768</math> possible hamburgers <math>\Rightarrow\mathrm{(C)}</math>.
== Video Solution by OmegaLearn ==
https://youtu.be/0W3VmFp55cM?t=373


==See Also==
~ pi_is_3.14


*[[2004 AMC 10A Problems]]
== Video Solution ==
https://youtu.be/3MiGotKnC_U?t=950


*[[2004 AMC 10A Problems/Problem 11|Previous Problem]]
~ ThePuzzlr


*[[2004 AMC 10A Problems/Problem 13|Next Problem]]
==Video Solution==
https://youtu.be/EIExyf8U7O0
 
Education, the Study of Everything
 
==Video Solution==
https://youtu.be/j-jNtSwTrxA?t=241 - AMBRIGGS
 
== See also ==
{{AMC10 box|year=2004|ab=A|num-b=11|num-a=13}}
 
[[Category:Introductory Combinatorics Problems]]
{{MAA Notice}}

Latest revision as of 22:27, 31 May 2025

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 a set with 8 elements. There are $8$ total condiments to choose from. Therefore, there are $2^8=256$ ways to order the condiments. (You have two choices for each condiment- one choice is to include that condiment, and the other choice is to not include that condiment.) 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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