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2004 AMC 10A Problem 12

2004 AMC 10A Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 2 of 5 · about 1 min · position P11-15

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?

A)
$24$
B)
$256$
C)
$768$
D)
$40,320$
E)
$120,960$

Answer

Key insight

Each of the 8 condiments is independently in or out, giving 2^8 subsets, times 3 patty choices.

Solution

A hamburger is specified by two independent decisions.

Condiments: for each of the $8$ condiments the customer decides yes or no, so there are $2^8 = 256$ possible collections (including the empty one).

Patties: $3$ options.

By the multiplication principle the number of hamburgers is $3 \cdot 256 = 768$ .

The answer is $\boxed{\textbf{(C)}\ 768}$ .

Why this works

"Any collection of" a set of $n$ items means any subset, and a set with $n$ elements has $2^n$ subsets because each element is independently included or excluded. Independent choices multiply. Distinguish this from arranging the condiments in order, which would be a factorial and is not what is being asked.

The trap

Using 8! = 40320 for the condiments as if their order mattered, choice (D).

Common mistakes

Using 8! = 40320 for the condiments as if their order mattered, choice (D).
Adding $3 + 256$ or forgetting the "no condiments" option (giving $3 \cdot 255$ ).

Techniques

Set up the equation/formula and compute; no special trick needed