AMC 8 · 2018 · #16

Grade 5 counting

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

permutations-basicfactorialcombinations-basic identify-subproblems ↑ Prerequisites: factorialmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

Problem

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?

Pick an answer.

(A)

1440

(B)

2880

(C)

5760

(D)

182,440

(E)

362,880

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

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.

1STEP 1

Glue each glued group into one super-book: Arabic → block A, Spanish → block S, leaving G₁ G₂ G₃ — now just 5 units to arrange.

{ A, S, G₁, G₂, G₃ } (5 units)

2STEP 2

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.

5! = 5 × 4 × 3 × 2 × 1 = 120

3STEP 3

Zoom inside block A: the two Arabic books swap as a₁a₂ or a₂a₁ — exactly 2! = 2 internal orders.

2! = 2

4STEP 4

Same zoom for block S: the four Spanish books line up in 4! = 4·3·2·1 = 24 internal orders.

4! = 4 × 3 × 2 × 1 = 24

5STEP 5

Multiply the three independent choices by the multiplication principle: 5! · 2! · 4! = 120 · 2 · 24 = 5,760 — answer (C).

5! × 2! × 4! = 120 × 2 × 24 = 5,760 → (C)

Answer

5760

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

\frac{1}{63}

of 9!. A sanity check on the ratio: glueing 2 books together reduces the count by a factor of

\frac{2!}{2 · (2-1)!}

\frac{1}{\frac{C(9, 2)}{8}}

… — simpler version: of all 9! orderings, the fraction with the Arabic pair adjacent is

\frac{2}{9}

and (independently, to first approximation) the fraction with the Spanish quartet adjacent is roughly

\frac{4! · 6!}{9!}

\frac{1}{21}

. Multiplying 9! ·

\frac{2}{9}

\frac{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.

💡Key takeaway

This AMC 8 problem only needs Grade 5 multiplication you already know — bundle the groups, count 5! × 2! × 4!, and you are done!

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