AMC 8 · 2001 · #14

Grade 4 counting

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

combinations-basicsystematic-enumerationmulti-digit-arithmetic identify-subproblemssystematic-enumeration ↑ Prerequisites: combinations-basic

📏 Medium solution 💡 3 insights

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Problem

Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose?

Meat: beef, chicken, pork
Vegetables: baked beans, corn, potatoes, tomatoes
Dessert: brownies, chocolate cake, chocolate pudding, ice cream

Pick an answer.

(A)

(B)

(C)

(D)

(E)

144

AMC 8 2001 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.

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Toolkit + CCSS Solution

Understand

Restated: Tyler builds a meal by picking one meat from $3$ choices, two different vegetables from $4$ choices, and one dessert from $4$ choices. The order of the items does not matter. How many different meals are possible?

Givens: Meats: beef, chicken, pork ($3$ options, pick $1$); Vegetables: baked beans, corn, potatoes, tomatoes ($4$ options, pick $2$ different); Desserts: brownies, chocolate cake, chocolate pudding, ice cream ($4$ options, pick $1$); Answer choices: (A) $4$, (B) $24$, (C) $72$, (D) $80$, (E) $144$

Unknowns: The total number of different meals Tyler can put together

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Make a Systematic List

A meal has three independent parts, so Tool #7 (Identify Subproblems) splits the count into three short pieces: pick the meat, pick the vegetable pair, pick the dessert. Multiply the three answers because the picks are independent. The only piece that needs care is the vegetable pair — "two different, order doesn't matter" is an unordered pair count. Tool #2 (Make a Systematic List) handles that without the $\binom{n}{k}$ formula: list each new pair exactly once. We avoid Tool #13 (Algebra) because plain multiplication of small whole numbers is enough.

Execute — Answer: C

#7 Identify Subproblems 3.OA.A.1 Step 1

Subproblem 1: count the meat choices.
Tyler picks one meat from a list of three.

$$\text{meat choices} = 3$$

💡 One pick out of three options is just three ways.

#2 Make a Systematic List 4.OA.A.3 Step 2

Subproblem 2: count the unordered pairs of vegetables.
Label the vegetables $1$ (baked beans), $2$ (corn), $3$ (potatoes), $4$ (tomatoes).
For each vegetable, list only the partners with a larger label so no pair is counted twice.

$$\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\} \;\Rightarrow\; 3 + 2 + 1 = 6 \text{ pairs}$$

💡 This is the same "handshake" count: vegetable $1$ has $3$ new partners, vegetable $2$ has $2$ new ones (its pair with $1$ is already counted), and so on.

#7 Identify Subproblems 3.OA.A.1 Step 3

Subproblem 3: count the dessert choices.
Tyler picks one dessert from a list of four.

$$\text{dessert choices} = 4$$

💡 One pick out of four options is four ways.

#7 Identify Subproblems 4.OA.A.3 Step 4

Combine the three subproblems.
The meat, vegetable pair, and dessert are picked independently, so the total number of meals is the product of the three counts.

$$3 \times 6 \times 4 = 72 \;\Rightarrow\; \textbf{(C)}$$

💡 Each meat can pair with each of the $6$ vegetable pairs ($3 \times 6 = 18$ meat-and-veg combos), and each of those can pair with each of the $4$ desserts ($18 \times 4 = 72$).

[1] #7 3.OA.A.1 Subproblem 1: count the meat choices. Tyler picks one meat from a list of three.

[2] #2 4.OA.A.3 Subproblem 2: count the unordered pairs of vegetables. Label the vegetables $1$

[3] #7 3.OA.A.1 Subproblem 3: count the dessert choices. Tyler picks one dessert from a list of

[4] #7 4.OA.A.3 Combine the three subproblems. The meat, vegetable pair, and dessert are picked

Review

Reasonableness: Size check: $72$ sits between $24$ (way too small, what you would get if the two vegetables counted as one choice of $4$) and $144$ (twice $72$, what you would get by treating the vegetable pair as ordered, $4 \times 3 = 12$, instead of $\tfrac{4 \times 3}{2} = 6$). The trap choices match common slips: (A) $4$ just adds $3 + 4 + 4 - 7$ style nonsense; (B) $24 = 3 \times 2 \times 4$ forgets the second vegetable; (D) $80$ is an off-pattern distractor; (E) $144 = 3 \times 12 \times 4$ double-counts each vegetable pair. Landing on (C) means each subproblem was counted correctly and the pairs were treated as unordered.

Alternative: Tool #9 (Solve an Easier Related Problem): drop the vegetable rule first. "$1$ meat, $1$ veg, $1$ dessert" gives $3 \times 4 \times 4 = 48$ meals. Now upgrade veg from "pick $1$" to "pick $2$ different (order doesn't matter)": that replaces the factor $4$ with $\tfrac{4 \times 3}{2} = 6$ (pick the first veg in $4$ ways, the second in $3$ ways, then divide by $2$ because the two orderings are the same meal). The new total is $3 \times 6 \times 4 = 72$, again (C).

CCSS standards used (min grade 4)

3.OA.A.1 Interpret products of whole numbers as the total number of objects in groups (Reading "$3$ meats" and "$4$ desserts" as the number of ways to make a single pick from each group.)
4.OA.A.3 Solve multistep word problems posed with whole numbers using the four operations (Listing the $6$ vegetable pairs systematically and multiplying $3 \times 6 \times 4 = 72$ to combine the three subproblem counts.)

⭐ Count each part of the meal on its own, then multiply. The only trick is the vegetable pair — list the pairs so none gets counted twice, and you get $6$, not $12$. $3 \times 6 \times 4 = 72$, answer (C).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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