Difference between revisions of "2004 AMC 10A Problems/Problem 12"
m (→Problem) |
(→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | + | 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> | |
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == | ||
Latest revision as of 12:03, 8 August 2024
Contents
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
Think of the condiments as in a set with 8 elements. 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. 
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
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. 
