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 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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

144

AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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 C(n, k) formula: list each new pair exactly once. We avoid Tool #13 (Algebra) because plain multiplication of small whole numbers is enough.

1STEP 1

Meat is one pick from three options, so there are 3 ways.

meat choices = 3

2STEP 2

The two vegetables are an unordered pair; listing each pair once gives 6 pairs.

{1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4} → 3 + 2 + 1 = 6 pairs

3STEP 3

Dessert is one pick from four options, so there are 4 ways.

dessert choices = 4

4STEP 4

The picks are independent, so multiply the three counts: 3 × 6 × 4 = 72, choice (C).

3 × 6 × 4 = 72 → (C)

Answer

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 × 3 = 12, instead of (4 × 3)/2 = 6). The trap choices match common slips: (A) 4 just adds 3 + 4 + 4 - 7 style nonsense; (B) 24 = 3 × 2 × 4 forgets the second vegetable; (D) 80 is an off-pattern distractor; (E) 144 = 3 × 12 × 4 double-counts each vegetable pair. Landing on (C) means each subproblem was counted correctly and the pairs were treated as unordered.

💡Key takeaway

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 × 6 × 4 = 72, answer (C).

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