AMC 8 · 2015 · #10
Grade 5 countingPick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
A 4-digit number is just four slots to fill — thousands, hundreds, tens, units. Tool #7 (Identify Subproblems) splits the count into four independent choices, one per slot, so the multiplication principle gives the total. To convince ourselves the multiplication is right, Tool #9 (Solve an Easier Related Problem) starts with a 2-digit version we can list by hand, then extends the pattern up to 4 digits.
Warm up on the 2-digit case: 9 tens-digit choices × 9 units-digit choices = 81, confirming the choices-times-choices logic scales up.
Solving the easier 2-digit case first shows that "choices per slot, then multiply" really works.
3.OA.A.1Solve An Easier Related ProblemSlot 1 (thousands): 0 is banned as the leading digit, so there are 9 choices, not 10.
Treating one digit position as its own mini-problem with its own constraint is the Tool #7 subproblems move.
4.OA.A.3Identify SubproblemsSlot 2 (hundreds): any digit except the thousands one (0 is fine here), so 10 - 1 = 9 choices.
Each slot's count depends on how many digits are still unused; subtract used digits from 10.
4.OA.A.3Identify SubproblemsSlot 3 (tens): must differ from the two digits already used, so 10 - 2 = 8 choices.
Same pattern: 10 digits minus those already used.
4.OA.A.3Identify SubproblemsSlot 4 (units): must differ from the three digits already used, so 10 - 3 = 7 choices.
Last slot has the fewest options because the most digits are already taken.
4.OA.A.3Identify SubproblemsMultiplication principle: multiply the four slot counts, 9 × 9 × 8 × 7 = 4536 → (B).
Each combination of slot choices gives a different number, and every valid number arises this way exactly once.
5.OA.A.1Identify SubproblemsBig counting problems get easy when you split them into one small choice per slot, then multiply — a Grade 5 expression-evaluation move.
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