Competition · AMC preparation · step 4 of 4
AMC 8 · 2001 · #14
Grade 4 countingPick an answer.
AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Count the meat choices
Meat is one pick from three options, so there are 3 ways.
One pick out of three options is just three ways.
3.OA.A.1Identify SubproblemsCount the vegetable pairs
The two vegetables are an unordered pair; listing each pair once gives 6 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.
Choosing two different vegetables from the four, when the order does not matter, can be done in exactly six ways.
▸ Why?
Fix an order for the four vegetables, then for each one count only the partners that come later in that order; this names every pair exactly once and sorts all of them into the tallies 3, 2, 1, 0, which add to 6.
▸ Why?
Each unordered pair lines up with exactly one "earlier vegetable, then later vegetable" listing — the single listing where its lower-ordered vegetable comes first — so the listings and the pairs match one for one, and no pair is skipped or counted twice.
▸ Why?
The pairs fall into separate groups by which vegetable is the earlier one — three start with the first vegetable, two with the second, one with the third, none with the last — and these groups do not overlap, so their sizes 3, 2, 1 add back to the full count.
Count the dessert choices
Dessert is one pick from four options, so there are 4 ways.
One pick out of four options is four ways.
3.OA.A.1Identify SubproblemsMultiply the three counts
The picks are independent, so multiply the three counts: 3 × 6 × 4 = 72, choice (C).
Each meat can pair with each of the 6 vegetable pairs (3 × 6 = 18 meat-and-veg combos), and each of those can pair with each of the 4 desserts (18 × 4 = 72).
4.OA.A.3Identify SubproblemsCount 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).
- Count the meat choices
- Count the vegetable pairs
- Count the dessert choices
- Multiply the three counts
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