Difference between revisions of "2004 AMC 10A Problems/Problem 12"
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| − | 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> | + | 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> |
== See also == | == See also == | ||
Revision as of 13:30, 7 April 2019
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?
Solution
For each condiment, a customer may either choose to order it or not. There are 
total condiments to choose from. 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 | ||
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