Difference between revisions of "2004 AMC 10A Problems/Problem 12"
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==Solution== | ==Solution== | ||
| − | For each condiment, | + | For each condiment, a customer may either order it or not. There are <math>8</math> condiments. 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>\ | ||
== See also == | == See also == | ||
Revision as of 23:38, 20 July 2014
Problem
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?
Solution
For each condiment, a customer may either order it or not. There are 
condiments. Therefore, there are 
ways to order the condiments. There are also 
choices for the meat, making a total of 
possible hamburgers. 
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. 
