Competition · AMC preparation · step 4 of 4
AMC 8 · 2018 · #16
Grade 5 countingPick an answer.
AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Counting all 9! arrangements and then filtering is hopeless by hand, so Tool #9 (Easier Related Problem) shrinks the picture: shrink each "must-stay-together" group to one super-book, count how many ways those super-books can be lined up, and a small concrete case (3 items: 3! = 6 orderings) makes the formula visible. Tool #7 (Identify Subproblems) then splits the full count into three clean pieces — outer arrangement of 5 units, internal arrangement of the Arabic block, internal arrangement of the Spanish block — and the multiplication principle combines them. Tool #2 (Systematic List) is the backup we lean on if the formula feels abstract: list the 2 orders of the Arabic pair and the 24 orders of the Spanish quartet to physically see the internal counts.
Bundle each group into one
Glue each glued group into one super-book: Arabic → block A, Spanish → block S, leaving G₁ G₂ G₃ — now just 5 units to arrange.
Bundling things that must travel together into one group is the same Grade 3 idea as "5 groups of 2" — we shrink many objects into a few groups so the count becomes manageable.
3.OA.A.1Solve An Easier Related ProblemOrder the 5 bundles
Arrange those 5 distinct units in a row: 5! = 5·4·3·2·1 = 120 orderings — the same logic as the 3! = 6 mini-case.
Multiplying 5 × 4 × 3 × 2 × 1 is Grade 5 multi-digit multiplication; the smaller 3! case shows why the answer is a product of descending choices.
5.NBT.B.5Solve An Easier Related ProblemOrder inside the Arabic block
Zoom inside block A: the two Arabic books swap as a₁a₂ or a₂a₁ — exactly 2! = 2 internal orders.
Listing the 2 internal orders of a pair is the most basic "2 groups of 1" arrangement count from Grade 3.
3.OA.A.1Make A Systematic ListOrder inside the Spanish block
Same zoom for block S: the four Spanish books line up in 4! = 4·3·2·1 = 24 internal orders.
Multiplying 4 × 3 × 2 × 1 is Grade 5 multi-digit multiplication; the systematic list shows why each new book multiplies the count by its own position choices.
5.NBT.B.5Make A Systematic ListMultiply the three counts
Multiply the three independent choices by the multiplication principle: 5! · 2! · 4! = 120 · 2 · 24 = 5,760 — answer (C).
Writing the total as the product 5! × 2! × 4! is Grade 5 expression-writing: it records the three independent choices as one calculation we can evaluate.
The number of valid shelf orderings equals the number of orderings of the 5 units, times the internal orderings of the Arabic block, times the internal orderings of the Spanish block.
▸ Why?
Every valid ordering is fixed by three choices made in turn — order the 5 units, then order the 2 books inside the Arabic block, then order the 4 books inside the Spanish block — and none of these choices limits the others.
▸ Why?
Each valid ordering matches exactly one triple (unit order, Arabic order, Spanish order), and each such triple rebuilds exactly one valid ordering, so counting the triples is the same as counting the orderings.
▸ Why?
The unit order, the Arabic order, and the Spanish order are independent choices — fixing one leaves every option open for the next — so the number of combined triples is the product of the three separate counts.
This AMC 8 problem only needs Grade 5 multiplication you already know — bundle the groups, count 5! × 2! × 4!, and you are done!
- Bundle each group into one
- Order the 5 bundles
- Order inside the Arabic block
- Order inside the Spanish block
- Multiply the three counts
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