AMC 8 · 2012 · #10
Grade 7 countingPick an answer.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are only a handful of arrangements of {0, 1, 2, 2}, so Tool #4 (Make a Systematic List) is the safest way to count without missing or double-counting — especially since the two 2s are identical. Tool #7 (Identify Subproblems) splits the list cleanly by the leading digit: either the number starts with 1 or it starts with 2 (it cannot start with 0). Counting each case separately and adding is much easier than juggling the whole thing at once.
A 4-digit number over 1000 needs a nonzero leading digit, and only 1 and 2 qualify — so split into two cases.
Knowing that the thousands place controls how big the number is — and that a 0 there shrinks it to 3 digits — is Grade 4 place-value reasoning.
4.NBT.A.2Identify SubproblemsCase 1 — lead with 1, then arrange {0, 2, 2} in the last three slots; the lone 0 has three spots, giving 3 numbers.
Writing out the three positions for the lone 0 is exactly the Grade 7 "organized list" method for counting outcomes.
7.SP.C.8Use Matrix LogicCase 2 — lead with 2, then arrange the distinct digits {0, 1, 2}; three different digits order 6 ways, giving 6 numbers.
Three distinct items have 3 × 2 × 1 = 6 orderings, and an organized list confirms every one.
7.SP.C.8Use Matrix LogicThe two cases can't overlap, so just add them: 3 + 6 = 9 numbers, which is (D).
Combining disjoint case counts by addition is the Grade 4 multi-step word-problem move.
4.OA.A.3Identify SubproblemsAn organized list — split by what the leading digit can be — turns this AMC 8 counting problem into a quick Grade 7 sample-space exercise.