AMC 10 · 2004 · #12
Grade 7 countingHenry'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?
Pick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: There are $8$ condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions.; The customer picks one, two, or three meat patties.; The customer may pick any collection of condiments, from none of them to all of them.; Answer choices: (A) $24$, (B) $256$, (C) $768$, (D) $40{,}320$, (E) $120{,}960$.
Unknowns: The total number of different hamburgers that can be ordered.
Understand
Restated: 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.
Givens: There are $8$ condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions.; The customer picks one, two, or three meat patties.; The customer may pick any collection of condiments, from none of them to all of them.; Answer choices: (A) $24$, (B) $256$, (C) $768$, (D) $40{,}320$, (E) $120{,}960$.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List
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.
Execute — Answer: C
7.SP.C.8 Step 1 Split into patties and condiments
- Building a hamburger means making two independent decisions: choosing the patty count and choosing the condiment collection.
- Because any patty count can be paired with any condiment collection, the total number of hamburgers is (number of patty options) $\times$ (number of condiment collections).
- So the whole problem reduces to counting those two pieces separately.
💡 When two choices don't affect each other, every option of one can be matched with every option of the other, so the counts multiply.
7.SP.C.8 Step 2 Count the patty options
- The customer chooses one, two, or three patties.
- That is a plain list of three possibilities, so there are exactly $3$ ways to pick the meat.
💡 "One, two, or three" is just three items on a list, so the count is three.
6.EE.A.1 Step 3 Count the condiment collections
- Handle each condiment on its own: for every one of the $8$ condiments, the customer makes an independent yes-or-no decision — put it on or leave it off.
- Each decision has $2$ outcomes, and there are $8$ of them made in a row, so by the multiplication counting principle the number of possible condiment collections is $2$ multiplied by itself $8$ times.
- That is $2^{8} = 256$.
- This automatically includes the plain burger with no condiments (all eight "no") and the loaded burger with all eight (all eight "yes").
💡 Each new condiment offers a fresh in-or-out switch, so every added condiment doubles the number of possible collections.
5.NBT.B.5 Step 4 Multiply the two counts
- Combine the pieces with the multiplication counting principle: $3$ patty options times $256$ condiment collections gives $3 \times 256 = 768$ different hamburgers.
- This is choice (C).
- Note the traps: $256$ alone (B) forgets the patties, and multiplying by the wrong factor like $8!$ (the arrangements of eight items) chases the large numbers $40{,}320$ (D) and $120{,}960$ (E), which mistake "any collection" for "an arrangement."
💡 Each of the three patty choices can wear any of the 256 condiment collections, so the collections repeat three times over.
7.SP.C.8 Building a hamburger means making two independent decisions: choosing the patty 7.SP.C.8 The customer chooses one, two, or three patties. That is a plain list of three p 6.EE.A.1 Handle each condiment on its own: for every one of the $8$ condiments, the custo 5.NBT.B.5 Combine the pieces with the multiplication counting principle: $3$ patty options Review
Reasonableness: 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 $120{,}960 = 3\times 8!$ 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.
Alternative: Instead of the power $2^8$, count condiment collections by size and add: $\binom{8}{0}+\binom{8}{1}+\cdots+\binom{8}{8} = 1+8+28+56+70+56+28+8+1 = 256$. The sum of a full row of Pascal's triangle is $2^8$, confirming $256$ collections. Multiplying by the $3$ patty options again gives $3\times 256 = 768$, choice (C).
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Recognizing the patty and condiment choices as independent decisions whose counts multiply, and counting each condiment as a separate yes/no outcome.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Turning eight independent two-way choices into the power $2^8$ and evaluating it as $256$.)5.NBT.B.5Fluently multiply multi-digit whole numbers using the standard algorithm (Multiplying the $3$ patty options by the $256$ condiment collections to get $768$.)
⭐ 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 $2^n$ combinations.
⭐ 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 $2^n$ combinations.
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