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

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There are also 3 choices for the meat, making a total of <math>256\times3=768</math> possible hamburgers <math>\Rightarrow\mathrm{(C)}</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>.
  
==See Also==
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== See also ==
 
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{{AMC10 box|year=2004|ab=A|num-b=11|num-a=13}}
*[[2004 AMC 10A Problems]]
 
 
 
*[[2004 AMC 10A Problems/Problem 11|Previous Problem]]
 
 
 
*[[2004 AMC 10A Problems/Problem 13|Next Problem]]
 
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]

Revision as of 02:40, 11 September 2007

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?

$\mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 256 \qquad \mathrm{(C) \ } 768 \qquad \mathrm{(D) \ } 40,320 \qquad \mathrm{(E) \ } 120,960$

Solution

For each condiment, our customer may either order it or not. There are 8 condiments. 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 $\Rightarrow\mathrm{(C)}$.

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 10 Problems and Solutions