Competition · AMC preparation · step 4 of 4

AMC 10 · 2019A · #17

Grade 7 geometry-3d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

permutations-basiccombinations-basiccombinatorial-identitycasework caseworkidentify-subproblems ↑ Prerequisites: permutations-basiccombinations-basic

📏 Short solution 💡 2 insights

Problem

A child has 2 red, 3 blue, and 4 green cubes — 9 cubes in total. She stacks 8 of them into a tower (one cube is left out). Cubes of the same color are indistinguishable. Count the number of different towers.

Pick an answer.

(A)

(B)

288

(C)

312

(D)

1,260

(E)

40,320

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): which cube color is left out is a clean way to split — three sub-cases give three smaller multinomial counts that we add. Tool #2 (Systematic List): inside each case the answer is a multinomial coefficient 8!/a! b! c!. Tool #3 (Eliminate): on a multiple-choice, sanity-check against the answer choices once one case is computed.

1STEP 1

Count the red leftover case

Case R (leave out one red): arrange 1 red, 3 blue, 4 green. The multinomial $\frac{8!}{1! 3! 4!}$ gives 280.

8!/1! 3! 4! = 40320/(1 · 6 · 24) = 40320/144 = 280

2STEP 2

Count the blue leftover case

Case B (leave out one blue): arrange 2 red, 2 blue, 4 green. Then $\frac{8!}{2! 2! 4!}$ gives 420.

8!/2! 2! 4! = 40320/(2 · 2 · 24) = 40320/96 = 420

3STEP 3

Count the green leftover case

Case G (leave out one green): arrange 2 red, 3 blue, 3 green. Then $\frac{8!}{2! 3! 3!}$ gives 560.

8!/2! 3! 3! = 40320/(2 · 6 · 6) = 40320/72 = 560

4STEP 4

Add the three cases

The three cases are disjoint, so add them: 280 + 420 + 560 = 1260, which is choice (D).

280 + 420 + 560 = 1260 → (D)

Answer

1,260

If all 9 cubes were distinct, there would be 9! = 362880 orderings — way more than any answer. Treating colors as repeats brings us down dramatically. A useful sanity check: pretending we use all 9 then deleting the last cube gives

\frac{9!}{2! 3! 4!}

\frac{362880}{288}

= 1260 — the same answer, by a one-to-one correspondence between 8-towers and 9-towers (the leftover is fixed to the bottom). Choice (E) 40320 = 8! is the all-distinct trap; (A) 24 and (B) 288 are far too small (they look like single denominators); (C) 312 is close to one case alone — all rule themselves out.

💡Key takeaway

This AMC 10 problem only needs Grade 7 organized-counting you already know — split by which cube is left out, compute one multinomial per case, add to get 280 + 420 + 560 = 1260. The answer is (D).

Count the red leftover case
Count the blue leftover case
Count the green leftover case
Add the three cases

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Count the red leftover case

Count the blue leftover case

Count the green leftover case

Add the three cases

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