AMC 10 · 2020 · #5
Grade 7 arithmeticProblem
How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
Pick an answer.
AMC 10 2020 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: Count the distinguishable ways to arrange in a row $1$ brown, $1$ purple, $2$ identical green, and $3$ identical yellow tiles — a total of $7$ tiles.
Givens: $1$ brown tile (unique); $1$ purple tile (unique); $2$ green tiles, indistinguishable from each other; $3$ yellow tiles, indistinguishable from each other; Tiles are placed in a row from left to right; Answer choices: (A) $210$, (B) $420$, (C) $630$, (D) $840$, (E) $1050$
Unknowns: Number of distinguishable arrangements of the $7$ tiles
Understand
Restated: Count the distinguishable ways to arrange in a row $1$ brown, $1$ purple, $2$ identical green, and $3$ identical yellow tiles — a total of $7$ tiles.
Givens: $1$ brown tile (unique); $1$ purple tile (unique); $2$ green tiles, indistinguishable from each other; $3$ yellow tiles, indistinguishable from each other; Tiles are placed in a row from left to right; Answer choices: (A) $210$, (B) $420$, (C) $630$, (D) $840$, (E) $1050$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem, #2 Make a Systematic List
Tool #7 (Subproblems): place the colors one at a time into the $7$ spots — first choose where the $3$ yellows go, then where the $2$ greens go, then the $2$ leftover spots get brown and purple. Multiplying the three sub-counts gives the total. Tool #9 (Easier Problem): teach the count-positions step on smaller boards first, e.g. "$2$ yellows among $3$ spots" gives $3$ patterns (YYS, YSY, SYY), to make $\binom{n}{k}$ concrete. Tool #2 (Systematic List): for each sub-count, list every arrangement in an orderly way (smallest position first) so nothing is missed.
Execute — Answer: B
7.SP.C.8 Step 1 - Subproblem 1 — place the $3$ yellow tiles in $3$ of the $7$ spots.
- Use Tool #9: try a smaller case first.
- Placing $2$ yellows in $3$ spots gives only $3$ patterns: positions $\{1,2\}, \{1,3\}, \{2,3\}$.
- So the count is "how many ways to pick the spots," not "how many orderings." For the real case, count subsets of size $3$ from $\{1,2,3,4,5,6,7\}$.
💡 On a small board you can list every pattern by hand; the rule that emerges is "pick the spots, don't order the tiles."
7.SP.C.8 Step 2 - Use a systematic list (or the small-case rule) to count $3$-spot subsets from $7$.
- Listing in order: $\{1,2,3\}, \{1,2,4\}, \dots, \{5,6,7\}$ gives $35$ subsets.
- (This is $\binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$.)
💡 Picking $3$ positions out of $7$ — easy to list in increasing order with no repeats.
7.SP.C.8 Step 3 - Subproblem 2 — after the yellows are placed, $4$ spots remain.
- Place $2$ greens in $2$ of those $4$ spots.
- By the same logic, the count is $\binom{4}{2} = \frac{4 \cdot 3}{2 \cdot 1} = 6$.
💡 Same subproblem, just with smaller numbers — pick $2$ of $4$ leftover spots.
4.OA.A.3 Step 4 - Subproblem 3 — $2$ spots remain.
- The brown and purple tiles are unique, so they can be placed in either order: $2 \times 1 = 2$ ways.
💡 Two distinct tiles into two spots is just "who goes first?" — $2$ choices.
4.OA.A.3 Step 5 Multiply the three independent choices (place yellows, then greens, then brown/purple): $35 \times 6 \times 2 = 420$.
💡 Each yellow layout pairs with every green layout pairs with every brown/purple swap — multiplication.
7.SP.C.8 Subproblem 1 — place the $3$ yellow tiles in $3$ of the $7$ spots. Use Tool #9: 7.SP.C.8 Use a systematic list (or the small-case rule) to count $3$-spot subsets from $7 7.SP.C.8 Subproblem 2 — after the yellows are placed, $4$ spots remain. Place $2$ greens 4.OA.A.3 Subproblem 3 — $2$ spots remain. The brown and purple tiles are unique, so they 4.OA.A.3 Multiply the three independent choices (place yellows, then greens, then brown/p Review
Reasonableness: Cross-check with the standard formula for distinguishable arrangements of $7$ items with repeats: $\frac{7!}{3! \cdot 2!} = \frac{5040}{12} = 420$. Same answer. Also $420$ sits in the middle of the choices, consistent with "not too many, not too few" arrangements when only $2$ tiles are unique.
Alternative: Tool #5 (Look for a Pattern) — compute $\frac{n!}{\text{(repeats)!}}$ directly: $\frac{7!}{1!\,1!\,2!\,3!} = \frac{5040}{12} = 420$. Faster once the formula is memorized; the subproblem approach above derives the formula.
CCSS standards used (min grade 7)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Multiplying the three independent sub-counts $35 \times 6 \times 2$ to combine the placement stages.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting ways to pick $3$ positions of $7$ (then $2$ of $4$) via organized listing of subsets.)
⭐ This AMC 10 problem only needs Grade 7 “organized lists / picking spots” you already know — pick $3$ yellow spots out of $7$ ($35$ ways), then $2$ green spots out of $4$ ($6$ ways), then $2$ ways for brown/purple, giving $35 \times 6 \times 2 = 420$.
⭐ This AMC 10 problem only needs Grade 7 “organized lists / picking spots” you already know — pick $3$ yellow spots out of $7$ ($35$ ways), then $2$ green spots out of $4$ ($6$ ways), then $2$ ways for brown/purple, giving $35 \times 6 \times 2 = 420$.
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