← Sensim Lab Sensim Math

AMC 10 · 2024 · #17

Grade 7 arithmetic
Archetype Casework by Position / Vertex Type
combinations-basicpermutations-basicset-partitionsystematic-enumeration caseworkidentify-subproblemssystematic-enumeration ↑ Prerequisites: combinations-basicpermutations-basicfactorial
📏 Long solution 💡 4 insights

Problem

In a race among 55 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?

(A) 180(B) 361(C) 420(D) 431(E) 720\textbf{(A) } 180 \qquad\textbf{(B) } 361 \qquad\textbf{(C) } 420 \qquad\textbf{(D) } 431 \qquad\textbf{(E) } 720

Pick an answer.

(A)
180
(B)
361
(C)
420
(D)
431
(E)
720
View mode:

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

#2 Make a Systematic List 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).
$$\text{Total} = \underbrace{N_0}_{k=0} + N_2 + N_3 + N_4 + N_5$$

💡 Casework with a small finite tag — here $k$ — is the cleanest Grade 5 "write the expression" setup.

#9 Solve an Easier Related Problem 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.
$$N_0 = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$

💡 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.

#9 Solve an Easier Related Problem 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.
$$N_2 = \binom{5}{2} \cdot 4! = 10 \cdot 24 = 240$$

💡 Group the tied snails together, count what's left as a smaller list-ordering problem — the Tool #9 "easier subproblem" move.

#9 Solve an Easier Related Problem 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.
$$N_3 = \binom{5}{3} \cdot 3! = 10 \cdot 6 = 60$$

💡 Same recipe, smaller: choose the $3$ tied, then line up $3$ blocks.

#9 Solve an Easier Related Problem 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.
$$N_4 = \binom{5}{4} \cdot 2! = 5 \cdot 2 = 10$$

💡 Two blocks left, so the only choice is which block goes first.

#9 Solve an Easier Related Problem 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.
$$N_5 = \binom{5}{5} \cdot 1! = 1 \cdot 1 = 1$$

💡 Total tie: one giant block, one possible outcome.

#7 Identify Subproblems 4.NBT.B.4 Step 7

Sum the five case counts.

$$\text{Total} = 120 + 240 + 60 + 10 + 1 = 431 \;\Rightarrow\; \textbf{(D)}$$

💡 Adding the five disjoint case counts is Grade 4 multi-digit addition, with no overlap to worry about.

[1] #2 5.OA.A.2 Set up the case split. Let $k$ be the tie size. The five cases are $k = 0$ (no t
[2] #9 4.OA.A.3 Case $k = 0$: no tie. All $5$ snails finish in different places, so this is just
[3] #9 7.SP.C.8 Case $k = 2$: exactly two snails tie. First pick which $2$ tie: $\binom{5}{2} =
[4] #9 7.SP.C.8 Case $k = 3$: three snails tie. Pick which $3$: $\binom{5}{3} = 10$. Then arrang
[5] #9 7.SP.C.8 Case $k = 4$: four snails tie. Pick which $4$: $\binom{5}{4} = 5$. Then $2$ bloc
[6] #9 7.SP.C.8 Case $k = 5$: all five tie. Pick which $5$: $\binom{5}{5} = 1$. Just $1$ block,
[7] #7 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.2 Write 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.3 Solve 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.8 Find 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.4 Fluently 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!

More like this

Same archetype — closest grade level first.