AMC 10 · 2004 · #12

Grade 7 counting
exponentssystematic-enumeration identify-subproblems ↑ Prerequisites: exponents
📏 Medium solution 💡 2 insights
Problem
A hamburger is built from two independent choices: the number of meat patties (one, two, or three) and any collection of the eight condiments. Count how many different hamburgers can be ordered.

Pick an answer.

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

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

A hamburger is really two decisions glued together: how many patties, and which condiments. Those decisions don't interfere with each other, so this is a signal to split the count into subproblems (Tool #7) — count the patty options, count the condiment collections, then multiply. The condiment part itself splits further: each of the 8 condiments is its own separate yes-or-no decision. Listing those choices systematically (Tool #2) shows that each condiment doubles the number of possible collections, which is exactly a power of two. The multiplication counting principle ties it together: when independent choices are made in sequence, the totals multiply.

1STEP 1

Split into patties and condiments

Two independent decisions build the burger: patty count and condiment collection. The total is (patty options) times (collections).

total = (patty options) × (condiment collections)
2STEP 2

Count the patty options

One, two, or three patties is just a three-item list, so there are exactly 3 ways to pick the meat.

patty options = 3
3STEP 3

Count the condiment collections

Each of the 8 condiments is an independent yes-or-no switch, so the collections number 2 to the eighth = 256, none and all included.

2⁸ = 2×2×…×2₈ = 256
4STEP 4

Multiply the two counts

3 patty options times 256 collections gives 768 hamburgers, choice (C) — 256 alone (B) forgets the patties.

3 × 256 = 768 → (C)
Answer
768
The answer 768 sits sensibly between the choices: it is exactly three times 256, matching the three patty options, so it must be a multiple of 256 larger than 256 itself — ruling out (A) 24 and (B) 256. The huge values 40,320 = 8! and 120,960 = 3 × 8! would require the condiments to be arranged in order, but the problem only asks which condiments are on the burger, not in what order, so those overcount badly. A moderate answer like 768 is exactly what a yes/no choice per condiment should give.
💡Key takeaway

When separate choices don't affect each other, count each one and multiply — and a yes-or-no switch on each of n items gives 2ⁿ combinations.

  • Split into patties and condiments
  • Count the patty options
  • Count the condiment collections
  • Multiply the two counts