AMC 8 · 2005 · #14
Grade 5 countingProblem
The Little Twelve Basketball Conference has two divisions, with six teams in each division. Each team plays each of the other teams in its own division twice and every team in the other division once. How many conference games are scheduled?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: The Little Twelve Basketball Conference has two divisions of $6$ teams each. Every team plays every other team in its own division twice, and every team in the other division once. How many conference games are scheduled in total?
Givens: Two divisions, $6$ teams each ($12$ teams total); Intra-division: each pair of teams in the same division plays $2$ games; Inter-division: each pair (one team from each division) plays $1$ game; Answer choices: (A) $80$, (B) $96$, (C) $100$, (D) $108$, (E) $192$
Unknowns: The total number of games scheduled across the whole conference
Understand
Restated: The Little Twelve Basketball Conference has two divisions of $6$ teams each. Every team plays every other team in its own division twice, and every team in the other division once. How many conference games are scheduled in total?
Givens: Two divisions, $6$ teams each ($12$ teams total); Intra-division: each pair of teams in the same division plays $2$ games; Inter-division: each pair (one team from each division) plays $1$ game; Answer choices: (A) $80$, (B) $96$, (C) $100$, (D) $108$, (E) $192$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List
The total schedule has two kinds of games with different rules, so Tool #7 (Identify Subproblems) splits the count into a clean sum: intra-division games (both divisions) plus inter-division games. Each subproblem is a short multiplication. To count the unordered team pairings inside one division, Tool #2 (Make a Systematic List) is the kid-friendly way to get $15$ without a combination formula — just list each team's new opponents in order so no pair gets counted twice. We avoid Tool #13 (Algebra) and the $\binom{n}{2}$ formula because grade-5 arithmetic is enough.
Execute — Answer: B
4.OA.A.3 Step 1 - Subproblem 1: count the unordered pairs of teams inside one division using a systematic list.
- Label the $6$ teams $1, 2, 3, 4, 5, 6$.
- For each team, list only the partners with a larger label so no pair is counted twice.
💡 Team $1$ has $5$ new partners, team $2$ has $4$ new ones (it already paired with team $1$), and so on. This is the same as the "handshake" count.
4.OA.A.1 Step 2 - Each of those $15$ pairs plays twice, and there are two divisions that follow the same rule.
- So the intra-division total multiplies the pairs-per-division by $2$ games and by $2$ divisions.
💡 "Plays twice" is a multiplicative comparison: it doubles the count. Two identical divisions double it again.
5.OA.A.2 Step 3 - Subproblem 2: count inter-division games.
- Each of the $6$ teams in Division A plays each of the $6$ teams in Division B exactly once, so this is a simple multiplication.
💡 Pairing every Division A team with every Division B team is a $6 \times 6$ grid of matchups.
4.OA.A.3 Step 4 Add the two subproblem totals to get the full schedule, then match to a choice.
💡 Intra-division and inter-division games never overlap, so the totals just add.
4.OA.A.3 Subproblem 1: count the unordered pairs of teams inside one division using a sys 4.OA.A.1 Each of those $15$ pairs plays twice, and there are two divisions that follow th 5.OA.A.2 Subproblem 2: count inter-division games. Each of the $6$ teams in Division A pl 4.OA.A.3 Add the two subproblem totals to get the full schedule, then match to a choice. Review
Reasonableness: Sanity check the size. A loose upper bound is to imagine every one of the $12$ teams playing every other team twice: that would be $12 \times 11 = 132$ ordered pairs, or $66$ unordered pairs, so at most $132$ games if every pair played twice. Our intra-division share ($60$) plus the once-only inter-division share ($36$) gives $96$, which sits comfortably below that bound. Trap choices come from common slips: $100$ is $60 + 40$ (over-counting inter-division), $108$ is $72 + 36$ (treating both intra and inter as $\times 2$), $192$ is $96 \times 2$ (counting each game from both teams' point of view), and $80$ is $40 + 40$ (forgetting the second division's intra games).
Alternative: Tool #9 (Solve an Easier Related Problem): shrink each division to $2$ teams. Intra-division: each division has $1$ pair playing twice, so $1 \times 2 \times 2 = 4$ games. Inter-division: $2 \times 2 = 4$ games. Total $8$. With $3$ teams each: intra $= 3 \times 2 \times 2 = 12$, inter $= 3 \times 3 = 9$, total $21$. The pattern "intra $= \binom{n}{2} \cdot 2 \cdot 2$ and inter $= n \cdot n$" holds; plugging $n = 6$ gives $60 + 36 = 96$, confirming (B).
CCSS standards used (min grade 5)
4.OA.A.1Interpret a multiplication equation as a comparison (Reading "each pair plays twice" as a multiplicative comparison that doubles the intra-division count.)4.OA.A.3Solve multistep word problems posed with whole numbers using the four operations (Listing the $15$ pairs per division and combining the intra- and inter-division subtotals into the final sum.)5.OA.A.2Write simple expressions that record calculations with numbers (Recording the inter-division count as $6 \times 6$ and the intra-division count as $15 \times 2 \times 2$ before evaluating.)
⭐ Split the schedule into "same-division" and "other-division" games, count each piece with a quick multiplication, then add. With that split, this AMC 8 problem becomes a Grade 5 multistep arithmetic exercise.
⭐ Split the schedule into "same-division" and "other-division" games, count each piece with a quick multiplication, then add. With that split, this AMC 8 problem becomes a Grade 5 multistep arithmetic exercise.
More like this
Same archetype — closest grade level first.
