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

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==Problem==
 
==Problem==
Henry's Hamburger Heaven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions.  A costomer can choose one, two, or three meat patties, and any collection of condiments.  How many different kinds of hamburgers can be ordered?
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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?
  
 
<math> \mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 256 \qquad \mathrm{(C) \ } 768 \qquad \mathrm{(D) \ } 40,320 \qquad \mathrm{(E) \ } 120,960  </math>
 
<math> \mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 256 \qquad \mathrm{(C) \ } 768 \qquad \mathrm{(D) \ } 40,320 \qquad \mathrm{(E) \ } 120,960  </math>
  
 
==Solution==
 
==Solution==
A costomer can either order a condiment, or not, and there are 8 condiments.  Therefore, there are <math>2^8=256</math> ways to order the condiments.
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For each condiment, our customer may either order it or not.  There are 8 condiments.  Therefore, there are <math>2^8=256</math> ways to order the condiments.
  
 
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>.

Revision as of 17:20, 12 November 2006

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