AMC 10 · 2019 · #17
Grade 7 geometry-3dProblem
A child builds towers using identically shaped cubes of different colors. How many different towers with a height 8 cubes can the child build with 2 red cubes, 3 blue cubes, and 4 green cubes? (One cube will be left out.)
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: $2$ red, $3$ blue, $4$ green — $9$ cubes total; Tower height $= 8$ cubes (so exactly $1$ cube is left out); Same-color cubes are interchangeable — only the color pattern matters; Choices: (A) $24$, (B) $288$, (C) $312$, (D) $1{,}260$, (E) $40{,}320$
Unknowns: Number of distinct color sequences of height $8$
Understand
Restated: 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.
Givens: $2$ red, $3$ blue, $4$ green — $9$ cubes total; Tower height $= 8$ cubes (so exactly $1$ cube is left out); Same-color cubes are interchangeable — only the color pattern matters; Choices: (A) $24$, (B) $288$, (C) $312$, (D) $1{,}260$, (E) $40{,}320$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
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 $\frac{8!}{a!\,b!\,c!}$. Tool #3 (Eliminate): on a multiple-choice, sanity-check against the answer choices once one case is computed.
Execute — Answer: D
7.SP.C.8 Step 1 - Split by which cube is left out.
- Case R (one red left out): use $1$ red, $3$ blue, $4$ green — $8$ cubes.
- Number of distinguishable orderings is the multinomial $\frac{8!}{1!\,3!\,4!}$.
💡 Indistinguishable cubes of the same color $\Rightarrow$ divide by each color's factorial.
7.SP.C.8 Step 2 - Case B (one blue left out): use $2$ red, $2$ blue, $4$ green — $8$ cubes.
- Count: $\frac{8!}{2!\,2!\,4!}$.
💡 Same recipe — only the color counts changed.
7.SP.C.8 Step 3 - Case G (one green left out): use $2$ red, $3$ blue, $3$ green — $8$ cubes.
- Count: $\frac{8!}{2!\,3!\,3!}$.
💡 Three cases, same multinomial recipe.
4.OA.A.3 Step 4 - The three cases are mutually exclusive (a single tower can have only one specific leftover-color profile), so add the counts: $280 + 420 + 560 = 1260$.
- That matches choice (D).
💡 Disjoint cases — just add. Quick choice-elimination confirms (D).
7.SP.C.8 Split by which cube is left out. Case R (one red left out): use $1$ red, $3$ blu 7.SP.C.8 Case B (one blue left out): use $2$ red, $2$ blue, $4$ green — $8$ cubes. Count: 7.SP.C.8 Case G (one green left out): use $2$ red, $3$ blue, $3$ green — $8$ cubes. Count 4.OA.A.3 The three cases are mutually exclusive (a single tower can have only one specifi Review
Reasonableness: 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.
Alternative: Tool #16 (Complement / clever bijection): every $8$-tower made from $\{2R, 3B, 4G\}$ corresponds to a unique $9$-cube arrangement (add the missing cube at the bottom). So the count is simply $\frac{9!}{2!\,3!\,4!} = 1260$ — one clean multinomial, no casework.
CCSS standards used (min grade 7)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Combining three case counts $280 + 420 + 560$ into the final total.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting distinguishable arrangements of $8$ items with repeated colors via the multinomial coefficient.)
⭐ 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 $\textbf{(D)}$.
⭐ 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 $\textbf{(D)}$.
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