2004 AMC 10A Problems/Problem 12

Revision as of 13:03, 8 August 2024 by Glowworm (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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?

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

Solution

Think of the condiments as in a set with 8 elements. There are $8$ total condiments to choose from. 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. $\boxed{\mathrm{(C)}\ 768}$

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

https://youtu.be/EIExyf8U7O0

Education, the Study of Everything

Video Solution

https://youtu.be/j-jNtSwTrxA?t=241 - AMBRIGGS

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png