AMC 10 · 2024 · #17
Grade 7 arithmeticProblem
In a race among 5 snails, there is at most one tie, but that tie can involve any number of snails. For example, the result might be that Dazzler is first; Abby, Cyrus, and Elroy are tied for second; and Bruna is fifth. How many different results of the race are possible?
Pick an answer.
AMC 10 2024 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: 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.
Givens: $5$ distinguishable snails; Either no tie at all, OR exactly one tie that groups $k$ snails for some $k \in \{2, 3, 4, 5\}$; All other snails (outside the tie) finish in distinct positions; Outcomes are distinct if either the assignment of snails to places, or the place ranks, differ; Answer choices: (A) $180$, (B) $361$, (C) $420$, (D) $431$, (E) $720$
Unknowns: Total count of distinct race outcomes
Understand
Restated: 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.
Givens: $5$ distinguishable snails; Either no tie at all, OR exactly one tie that groups $k$ snails for some $k \in \{2, 3, 4, 5\}$; All other snails (outside the tie) finish in distinct positions; Outcomes are distinct if either the assignment of snails to places, or the place ranks, differ; Answer choices: (A) $180$, (B) $361$, (C) $420$, (D) $431$, (E) $720$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #9 Solve an Easier Related Problem
"At most one tie" is the big switch — it splits the problem into five clean cases by tie size $k \in \{0, 2, 3, 4, 5\}$ (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 $\binom{5}{k}$ picks which snails are in the tie.
Execute — Answer: D
5.OA.A.2 Step 1 - Set up the case split.
- Let $k$ be the tie size.
- The five cases are $k = 0$ (no tie) and $k \in \{2, 3, 4, 5\}$ (one tie of size $k$).
- For each case, the count of outcomes is (ways to pick which snails tie) $\times$ (ways to rank the resulting blocks).
💡 Casework with a small finite tag — here $k$ — is the cleanest Grade 5 "write the expression" setup.
4.OA.A.3 Step 2 - Case $k = 0$: no tie.
- All $5$ snails finish in different places, so this is just the number of orderings of $5$ distinct items.
💡 Grade 4 multi-step arithmetic: $5 \times 4 \times 3 \times 2 \times 1 = 120$, the count of ways to put $5$ different things in a row.
7.SP.C.8 Step 3 - Case $k = 2$: exactly two snails tie.
- First pick which $2$ tie: $\binom{5}{2} = 10$.
- Treat the tied pair as a single block; now we have $4$ blocks (the pair + $3$ singletons) to arrange in $4! = 24$ strict orders.
💡 Group the tied snails together, count what's left as a smaller list-ordering problem — the Tool #9 "easier subproblem" move.
7.SP.C.8 Step 4 - Case $k = 3$: three snails tie.
- Pick which $3$: $\binom{5}{3} = 10$.
- Then arrange $3$ blocks (the trio + $2$ singletons): $3! = 6$ ways.
💡 Same recipe, smaller: choose the $3$ tied, then line up $3$ blocks.
7.SP.C.8 Step 5 - Case $k = 4$: four snails tie.
- Pick which $4$: $\binom{5}{4} = 5$.
- Then $2$ blocks (the quartet + $1$ singleton) line up in $2! = 2$ ways.
💡 Two blocks left, so the only choice is which block goes first.
7.SP.C.8 Step 6 - Case $k = 5$: all five tie.
- Pick which $5$: $\binom{5}{5} = 1$.
- Just $1$ block, $1! = 1$ ordering.
💡 Total tie: one giant block, one possible outcome.
4.NBT.B.4 Step 7 Sum the five case counts.
💡 Adding the five disjoint case counts is Grade 4 multi-digit addition, with no overlap to worry about.
5.OA.A.2 Set up the case split. Let $k$ be the tie size. The five cases are $k = 0$ (no t 4.OA.A.3 Case $k = 0$: no tie. All $5$ snails finish in different places, so this is just 7.SP.C.8 Case $k = 2$: exactly two snails tie. First pick which $2$ tie: $\binom{5}{2} = 7.SP.C.8 Case $k = 3$: three snails tie. Pick which $3$: $\binom{5}{3} = 10$. Then arrang 7.SP.C.8 Case $k = 4$: four snails tie. Pick which $4$: $\binom{5}{4} = 5$. Then $2$ bloc 7.SP.C.8 Case $k = 5$: all five tie. Pick which $5$: $\binom{5}{5} = 1$. Just $1$ block, 4.NBT.B.4 Sum the five case counts. Review
Reasonableness: 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 $5 \cdot 2 = 10$ — 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.
Alternative: Tool #5 (Look for a Pattern) — the count of "ordered partitions of $n$ snails into at most $2$ ranks, OR no tie" can also be read as $a(n) = n! + \sum_{k=2}^{n} \binom{n}{k}(n-k+1)!$. For $n = 5$, this expansion gives $120 + 240 + 60 + 10 + 1 = 431$ as before. The pattern view is reassuring but the casework above is faster.
CCSS standards used (min grade 7)
5.OA.A.2Write simple expressions that record calculations with numbers (Writing the total as the sum $N_0 + N_2 + N_3 + N_4 + N_5$ of five disjoint cases by tie size.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Computing $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ for the no-tie case.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting choices via $\binom{5}{k}$ (which snails tie) times $(6-k)!$ (block orderings) for each tie size $k$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the five case totals $120 + 240 + 60 + 10 + 1 = 431$.)
⭐ 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!
⭐ 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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