Art of Problem Solving

2004 AMC 10A Problems/Problem 12: Difference between revisions

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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?
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.
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