AMC 10 · 2022 · #14
Grade 7 countingProblem
How many ways are there to split the integers through into pairs such that in each pair, the greater number is at least times the lesser number?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Take the integers $1, 2, 3, \ldots, 14$ and split them into $7$ pairs so that in every pair the bigger number is at least twice the smaller one. How many such pairings are there?
Givens: Numbers $1$ through $14$, each used exactly once; $7$ unordered pairs in total; In every pair $(a, b)$ with $a < b$, we require $b \ge 2a$; Answer choices: (A) $108$, (B) $120$, (C) $126$, (D) $132$, (E) $144$
Unknowns: Number of valid ways to form the $7$ pairs
Understand
Restated: Take the integers $1, 2, 3, \ldots, 14$ and split them into $7$ pairs so that in every pair the bigger number is at least twice the smaller one. How many such pairings are there?
Givens: Numbers $1$ through $14$, each used exactly once; $7$ unordered pairs in total; In every pair $(a, b)$ with $a < b$, we require $b \ge 2a$; Answer choices: (A) $108$, (B) $120$, (C) $126$, (D) $132$, (E) $144$
Plan
Primary tool: #9 Easier Related Problem
Secondary: #2 Make a Systematic List, #16 Change Focus / Complement, #5 Look for a Pattern, #3 Eliminate Possibilities
Tool #9 (Easier Related Problem) re-frames the question: instead of "pair $14$ numbers with a $2\times$ rule", first ask the easier structural question "can two big numbers ever be paired?". The answer (no — see Step 1) collapses the problem to a bijection between $L = \{1,\ldots,7\}$ and $G = \{8,\ldots,14\}$, halving the search space. Tool #2 (Systematic List) then tabulates, for each $a \in L$, the set of legal partners in $G$. Tool #5 (Pattern) handles the order: assign partners starting from the most constrained $a$ (which has the fewest legal $b$'s) and work down. The multiplication principle multiplies the choice counts. Tool #16 (Change Focus) is a sanity-check angle: the smaller $a$'s ($a \le 4$) are completely unconstrained at the end, so they contribute $4!$ in one chunk. Tool #3 closes by matching the final integer to the answer list.
Execute — Answer: E
6.NS.C.7 Step 1 - Tool #9 (Easier Related Problem): first answer the easier structural question — can both members of a pair come from $G = \{8, 9, 10, 11, 12, 13, 14\}$?
- For any $x < y$ in $G$, the smallest gap is $(8, 9)$, but $9 < 16 = 2 \times 8$.
- More generally if $x \ge 8$ then $2x \ge 16 > y$ for every $y \le 14$.
- So NO pair can sit entirely inside $G$.
- Each of the $7$ numbers in $G$ must therefore be the LARGER member of one pair, and each of $L = \{1, \ldots, 7\}$ must be the SMALLER.
- The problem reduces to a bijection $f : L \to G$ with $f(a) \ge 2a$.
💡 Grade 6 number-ordering: comparing $2x$ with the largest possible $y$ rules out any in-$G$ pair without trying examples.
6.NS.C.7 Step 2 - Tool #2 (Systematic List): for each $a \in L$, list the elements of $G$ that satisfy $b \ge 2a$.
- The smaller $a$ gets, the more partners it has.
- The number $7$ is the tightest: only $14$ works.
💡 Listing partners row-by-row makes the constraint structure visible at a glance — the tail of $L$ is where the bottleneck lives.
7.SP.C.8 Step 3 - Tool #5 (Pattern): assign partners from the most constrained $a$ down to the least.
- Start at $a = 7$ (only choice $14$), then $a = 6$ ($\{12, 13\}$ left, $2$ choices), then $a = 5$.
- For $a = 5$ the legal set originally was $\{10, 11, 12, 13\}$ minus the one used by $6$ and minus $14$ (used by $7$) — so $5 - 2 = 3$ choices regardless of which one $6$ picked.
💡 Hardest-first ordering means each later count is unambiguous — Grade 7 organized-listing of compound events.
7.SP.C.8 Step 4 - After $7, 6, 5$ are assigned, three numbers from $G$ are used and four remain in some four-element subset $G''$ of $\{8, 9, 10, 11, 12, 13\}$.
- Crucially every $b \in G''$ satisfies $b \ge 8$.
- For the four leftover smalls $a \in \{1, 2, 3, 4\}$, the binding constraint is $a = 4$ requiring $b \ge 8$ — automatically satisfied by every $b \in G''$.
- So $\{1, 2, 3, 4\}$ matches freely with $G''$ in $4! = 24$ ways.
💡 Tool #16 angle: instead of tracking individual constraints, notice they're ALL automatically met — every remaining $b \ge 8 \ge 2 \cdot 4$, so freedom is total.
7.SP.C.8 Step 5 Multiply the independent stages together (multiplication principle).
💡 Grade 7 fundamental counting: multiply choices across independent stages.
6.EE.B.5 Step 6 Match $144$ to the choices: option (E).
💡 Final compare to the multiple-choice list.
6.NS.C.7 Tool #9 (Easier Related Problem): first answer the easier structural question — 6.NS.C.7 Tool #2 (Systematic List): for each $a \in L$, list the elements of $G$ that sat 7.SP.C.8 Tool #5 (Pattern): assign partners from the most constrained $a$ down to the lea 7.SP.C.8 After $7, 6, 5$ are assigned, three numbers from $G$ are used and four remain in 7.SP.C.8 Multiply the independent stages together (multiplication principle). 6.EE.B.5 Match $144$ to the choices: option (E). Review
Reasonableness: Quick magnitude check. A loose upper bound on the count is $7!$ (any bijection between $L$ and $G$), which is $5040$. The constraint $b \ge 2a$ should cut that down significantly — and indeed $144$ is about $7!/35$, much smaller. Also $144 = 1 \times 2 \times 3 \times 4!$ has a clean combinatorial story: three forced/restricted choices on the tail of $L$ and a free $4!$ on the head. Spot-check by enumerating $a = 7$'s and $a = 6$'s choices: $7$ must take $14$, $6$ has $\{12, 13\}$ (two cases). In each of those two cases, $5$ then has exactly three legal partners (from $\{10, 11, 12, 13\}$ minus the one $6$ took), giving $2 \times 3 = 6$ ways for the top three, times $4! = 24$ for the bottom four = $144$. ✓
Alternative: Tool #11 (Work Backwards) — set up the constraints in the opposite order, from $a = 1$ down to $a = 7$. The problem is harder this way because $a = 1$'s seven choices interact with everyone else. So we'd track an evolving "forbidden" set per step. This confirms the wisdom of the hardest-first ordering Tool #5 chose. Either way, by direct enumeration (or recursive program) the count is $144$.
CCSS standards used (min grade 7)
6.EE.B.5Understand solving an equation or inequality as a process of finding values (Comparing the computed $144$ to the five answer choices and selecting option (E).)6.NS.C.7Understand ordering and absolute value of rational numbers (Comparing $2x$ to $y$ for $x, y$ in $G$ to rule out any in-$G$ pair, and listing legal partners for each $a \in L$ by ordering.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the independent stages of the constrained pairing as an organized list, using the multiplication principle $1 \cdot 2 \cdot 3 \cdot 4!$ for compound events.)
⭐ This AMC 10 problem only needs Grade 7 organized-list counting you already know — once you notice that the big numbers $8$-$14$ can never pair with each other, the problem becomes a clean bijection. Pair the hardest first ($7$ with $14$, then $6$, then $5$), and the four leftover small numbers are free, giving $1 \times 2 \times 3 \times 4! = 144$.
⭐ This AMC 10 problem only needs Grade 7 organized-list counting you already know — once you notice that the big numbers $8$-$14$ can never pair with each other, the problem becomes a clean bijection. Pair the hardest first ($7$ with $14$, then $6$, then $5$), and the four leftover small numbers are free, giving $1 \times 2 \times 3 \times 4! = 144$.
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