AMC 8 · 2018 · #16
Easy mode Grade 5Problem
Imagine a shelf with 9 books standing in a row. They all look different.
2 of the books are in Arabic. 3 are in German. 4 are in Spanish.
Professor Chang wants to line them up with two rules:
- The 2 Arabic books must sit right next to each other, with no other book between them.
- The 4 Spanish books must also sit all together in a row, with no other book between them.
The 3 German books can go anywhere.
How many different ways can Professor Chang line up the 9 books?
Pick an answer.
AMC 8 2018 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: Professor Chang lines up $9$ distinct language books on a single shelf: $2$ Arabic, $3$ German, and $4$ Spanish. The two Arabic books must stand next to each other (forming one contiguous block) and the four Spanish books must stand next to each other (another contiguous block). The three German books have no adjacency requirement. How many shelf orderings satisfy these two togetherness rules?
Givens: $9$ distinct books in total; $2$ Arabic books that must be adjacent; $4$ Spanish books that must be adjacent; $3$ German books with no adjacency constraint; Answer choices: (A) $1{,}440$, (B) $2{,}880$, (C) $5{,}760$, (D) $182{,}440$, (E) $362{,}880$
Unknowns: The number of valid shelf orderings of all $9$ books that respect both togetherness constraints
Understand
Restated: Professor Chang lines up $9$ distinct language books on a single shelf: $2$ Arabic, $3$ German, and $4$ Spanish. The two Arabic books must stand next to each other (forming one contiguous block) and the four Spanish books must stand next to each other (another contiguous block). The three German books have no adjacency requirement. How many shelf orderings satisfy these two togetherness rules?
Givens: $9$ distinct books in total; $2$ Arabic books that must be adjacent; $4$ Spanish books that must be adjacent; $3$ German books with no adjacency constraint; Answer choices: (A) $1{,}440$, (B) $2{,}880$, (C) $5{,}760$, (D) $182{,}440$, (E) $362{,}880$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
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.
Execute — Answer: C
3.OA.A.1 Step 1 - Treat each must-stay-together group as a single super-book.
- The $2$ Arabic books become one block $\mathbf{A}$, the $4$ Spanish books become one block $\mathbf{S}$, and the $3$ German books $G_1, G_2, G_3$ stay individual.
- Now we are only arranging $5$ items on the shelf, which is an enormous simplification.
💡 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.
5.NBT.B.5 Step 2 - Count the orderings of the $5$ super-units.
- We can verify the pattern with a Tool #9 mini-case: $3$ distinct items have $3! = 6$ orderings (ABC, ACB, BAC, BCA, CAB, CBA).
- The same logic gives $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$ orderings of our $5$ units.
💡 Multiplying $5 \times 4 \times 3 \times 2 \times 1$ is Grade 5 multi-digit multiplication; the smaller $3!$ case shows why the answer is a product of descending choices.
3.OA.A.1 Step 3 - Now zoom into the Arabic block.
- Inside that block the two Arabic books $a_1, a_2$ can be ordered as $a_1 a_2$ or $a_2 a_1$ — exactly $2! = 2$ ways (Tool #2 confirms by listing).
💡 Listing the $2$ internal orders of a pair is the most basic "$2$ groups of $1$" arrangement count from Grade 3.
5.NBT.B.5 Step 4 - Do the same zoom for the Spanish block.
- Four distinguishable books $s_1, s_2, s_3, s_4$ can be ordered internally in $4! = 24$ ways.
- A Tool #2 systematic listing (fix $s_1$ first, then $s_2$, etc.) confirms there are $4 \cdot 3 \cdot 2 \cdot 1 = 24$ orderings.
💡 Multiplying $4 \times 3 \times 2 \times 1$ is Grade 5 multi-digit multiplication; the systematic list shows why each new book multiplies the count by its own position choices.
5.OA.A.2 Step 5 - Combine the three independent choices using the multiplication principle: outer arrangement of the $5$ units, times internal arrangement of the Arabic block, times internal arrangement of the Spanish block.
- This matches the answer choice (C).
💡 Writing the total as the product $5! \times 2! \times 4!$ is Grade 5 expression-writing: it records the three independent choices as one calculation we can evaluate.
3.OA.A.1 Treat each must-stay-together group as a single super-book. The $2$ Arabic books 5.NBT.B.5 Count the orderings of the $5$ super-units. We can verify the pattern with a Too 3.OA.A.1 Now zoom into the Arabic block. Inside that block the two Arabic books $a_1, a_2 5.NBT.B.5 Do the same zoom for the Spanish block. Four distinguishable books $s_1, s_2, s_ 5.OA.A.2 Combine the three independent choices using the multiplication principle: outer Review
Reasonableness: The unrestricted count of arranging $9$ distinct books is $9! = 362{,}880$ (choice (E)). Our togetherness rules should *cut down* this number, and $5{,}760$ is indeed far smaller — about $\tfrac{1}{63}$ of $9!$. A sanity check on the ratio: glueing $2$ books together reduces the count by a factor of $\tfrac{2!}{2 \cdot (2-1)!} \cdot \tfrac{1}{\binom{9}{2}/8} \dots$ — simpler version: of all $9!$ orderings, the fraction with the Arabic pair adjacent is $\tfrac{2}{9}$ and (independently, to first approximation) the fraction with the Spanish quartet adjacent is roughly $\tfrac{4! \cdot 6!}{9!} = \tfrac{1}{21}$. Multiplying $9! \cdot \tfrac{2}{9} \cdot \tfrac{1}{21}$ gives about $3{,}840$, the same order of magnitude as $5{,}760$ — comforting agreement. Choices (D) $182{,}440$ and (E) $362{,}880$ would mean the constraints didn't tighten the count enough, so they fail the reasonableness test.
Alternative: Tool #3 (Eliminate Possibilities) prunes the answer list using parity and divisibility. The total must be divisible by $5! = 120$ (the outer arrangements), so any choice not divisible by $120$ is out: (D) $182{,}440 / 120 = 1520.33\ldots$ — eliminated. The total must also be divisible by $4! = 24$ for the internal Spanish orderings; (A) $1{,}440 = 60 \cdot 24$ checks, (B) $2{,}880 = 120 \cdot 24$ checks, (C) $5{,}760 = 240 \cdot 24$ checks, (E) $362{,}880 = 9!$ would require no togetherness reduction at all. Then noticing that we need both the $\times 2$ (Arabic) and $\times 24$ (Spanish) factors *on top of* $5! = 120$ gives $120 \times 2 \times 24 = 5{,}760$, picking (C).
CCSS standards used (min grade 5)
3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Interpreting the Arabic pair and the Spanish quartet as "groups" of books we bundle together, and reading internal orderings ($2!$, $4!$) as "total objects in groups" counts.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Computing the factorials and the final product: $5! = 120$, $4! = 24$, and $120 \times 2 \times 24 = 5{,}760$ all rely on Grade 5 fluent multi-digit multiplication.)5.OA.A.2Write simple expressions that record calculations with numbers (Recording the multiplication principle as the single expression $5! \times 2! \times 4!$, which captures all three independent choices in one calculation.)
⭐ This AMC 8 problem only needs Grade 5 multiplication you already know — bundle the groups, count $5! \times 2! \times 4!$, and you are done!
⭐ This AMC 8 problem only needs Grade 5 multiplication you already know — bundle the groups, count $5! \times 2! \times 4!$, and you are done!
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