AMC 10 · 2022 · #14
Grade 7 countingProblem
How many ways are there to split the integers 1 through 14 into 7 pairs such that in each pair, the greater number is at least 2 times the lesser number?
Pick an answer.
AMC 10 2022 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: 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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