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

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==Solution==
 
==Solution==
 
For each condiment, a customer may either choose to order it or not.  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>
 
For each condiment, a customer may either choose to order it or not.  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>
 +
 +
== Video Solution ==
 +
https://youtu.be/3MiGotKnC_U?t=950
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~ ThePuzzlr
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 09:09, 19 October 2021

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

For each condiment, a customer may either choose to order it or not. 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

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

~ ThePuzzlr

Video Solution

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

~ pi_is_3.14

Video Solution

https://youtu.be/EIExyf8U7O0

Education, the Study of Everything

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

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