AMC 10 · 2024 · #17

Grade 7 arithmetic

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

combinations-basicpermutations-basicset-partitionsystematic-enumeration caseworkidentify-subproblemssystematic-enumeration ↑ Prerequisites: combinations-basicpermutations-basicfactorial

📏 Long solution 💡 4 insights

Problem

Five named snails finish a race. At most one tie is allowed, but a tie can group any 2, 3, 4, or all 5 snails into one place. Count every possible distinct ordering of finishes.

Pick an answer.

(A)

180

(B)

361

(C)

420

(D)

431

(E)

720

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

"At most one tie" is the big switch — it splits the problem into five clean cases by tie size k ∈ {0, 2, 3, 4, 5} (tie size 1 is just "no tie"). That fits Tool #7 (Identify Subproblems) perfectly: solve each case, then add. Tool #2 (Make a Systematic List) gives the case ordering and prevents double-counting. Inside each case we use Tool #9 (Solve an Easier Related Problem): tie k snails, treat them as one block, and the leftover problem is just "line up 6 - k blocks in a strict order" — a tiny factorial. The factor C(5, k) picks which snails are in the tie.

1STEP 1

Split by tie size k: cases are k = 0 (no tie) and k ∈ {2, 3, 4, 5}; each count is (who ties) × (how the blocks rank).

Total = N₀_k=0 + N₂ + N₃ + N₄ + N₅

2STEP 2

Case k = 0: all 5 finish in distinct places, so 5! = 120 orderings.

N₀ = 5! = 5 × 4 × 3 × 2 × 1 = 120

3STEP 3

Case k = 2: choose the tied pair C(5,2) = 10, then rank 4 blocks 4! = 24, giving 240.

N₂ = C(5, 2) · 4! = 10 · 24 = 240

4STEP 4

Case k = 3: choose 3 tied C(5,3) = 10, then rank 3 blocks 3! = 6, giving 60.

N₃ = C(5, 3) · 3! = 10 · 6 = 60

5STEP 5

Case k = 4: choose 4 tied C(5,4) = 5, then rank 2 blocks 2! = 2, giving 10.

N₄ = C(5, 4) · 2! = 5 · 2 = 10

6STEP 6

Case k = 5: all tie, C(5,5) = 1 and 1! = 1, giving 1 outcome.

N₅ = C(5, 5) · 1! = 1 · 1 = 1

7STEP 7

Add the five disjoint case counts: 120 + 240 + 60 + 10 + 1 = 431.

Total = 120 + 240 + 60 + 10 + 1 = 431 → (D)

Answer

431

Tie-size cases are mutually exclusive by construction (a single race has one tie size), so the sum rule applies without subtraction. Spot-check the small case k = 5 by hand: only 1 outcome (everyone first). Spot-check k = 4: pick the lone non-tied snail in 5 ways, then it is either ahead of or behind the tied quartet, giving 5 · 2 = 10 — matches. Total 431 is choice (D). For an upper-bound sanity check: with no tie rule at all, the count of ordered 5-tuples is 5! = 120 orderings, but adding tied outcomes inflates this — 431 sits well below the 720 of (E) (a 5!-ish overcount), which is what one would expect.

💡Key takeaway

This AMC 10 problem only needs Grade 7 organized counting — list each tie-size case, multiply (who ties) by (how the blocks line up), and add — that you already know!

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