AMC 10 · 2020 · #5
Grade 7 arithmeticPick an answer.
AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 C(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.
Subproblem 1: place the 3 yellows. A smaller case (2 yellows in 3 spots gives 3 patterns) shows you pick the spots, not order the tiles.
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.8Solve An Easier Related ProblemList the size-3 subsets of {1,…,7} in order — {1,2,3}, …, {5,6,7} — giving 35, which is C(7, 3).
Picking 3 positions out of 7 — easy to list in increasing order with no repeats.
7.SP.C.8Make A Systematic ListSubproblem 2: 4 spots remain — choosing 2 for the greens is C(4, 2) = 6.
Same subproblem, just with smaller numbers — pick 2 of 4 leftover spots.
7.SP.C.8Identify SubproblemsSubproblem 3: 2 spots left — the unique brown and purple can go in either order, so 2 ways.
Two distinct tiles into two spots is just "who goes first?" — 2 choices.
4.OA.A.3Identify SubproblemsMultiply the three independent counts: 35 × 6 × 2 = 420.
Each yellow layout pairs with every green layout pairs with every brown/purple swap — multiplication.
4.OA.A.3Identify SubproblemsThis 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 × 6 × 2 = 420.