AMC 8 · 2001 · #14
Easy mode Grade 4Problem
Tyler is going through a buffet line. He builds one meal by picking:
- 1 kind of meat, from: beef, chicken, pork
- 2 different vegetables, from: baked beans, corn, potatoes, tomatoes
- 1 dessert, from: brownies, chocolate cake, chocolate pudding, ice cream
For the two vegetables, the order doesn't matter. Corn-and-potatoes counts the same as potatoes-and-corn.
How many different meals can Tyler put together?
Pick an answer.
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
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) $144$
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
3.OA.A.1 Step 1 - Subproblem 1: count the meat choices.
- Tyler picks one meat from a list of three.
💡 One pick out of three options is just three ways.
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.
💡 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.
3.OA.A.1 Step 3 - Subproblem 3: count the dessert choices.
- Tyler picks one dessert from a list of four.
💡 One pick out of four options is four ways.
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.
💡 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 ($18 \times 4 = 72$).
3.OA.A.1 Subproblem 1: count the meat choices. Tyler picks one meat from a list of three. 4.OA.A.3 Subproblem 2: count the unordered pairs of vegetables. Label the vegetables $1$ 3.OA.A.1 Subproblem 3: count the dessert choices. Tyler picks one dessert from a list of 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.1Interpret 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.3Solve 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).
⭐ 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).
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