2004 AMC 10A Problem 12

P11-15~1 minPrint

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

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?

Show solution without answering

Key insight

For each of the 8 condiments you just say yes or no, giving 2 to the 8th = 256 choices. Multiply by 3 patty choices.

Solution

A hamburger is decided by two separate choices: the condiments and the number of patties.

1. Condiments. Go through the $8$ condiments one at a time and decide yes or no for each. That is $2$ choices, $8$ times:
$2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^8 = 256.$
This includes the plain burger with no condiments, which "any collection" allows.
2. Patties. One, two or three: $3$ choices.
3. Every condiment choice can go with every patty choice, so multiply:
$3 \times 256 = 768.$

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

Why this works

"Any collection" of $n$ items means you decide in or out for each one, which gives $2^n$ possibilities. Separate choices multiply. The order in which condiments go on the burger is not being asked about, so factorials do not belong here.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(C)}\ 768}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
Introduction to Counting & Probability, Chapter 4: Committees and Combinations