2022 AMC 10A Problem 14

A number $x$ can be the smaller member of a pair only if $2x \le 14$, so $8, 9, \dots, 14$ must all be the larger members. That is seven numbers, and there are exactly seven pairs, so the larger members are precisely $8$ through $14$ and the smaller members are precisely $1$ through $7$.

Each of $1, \dots, 7$ must be matched with a number from $8, \dots, 14$ at least twice as large. Match the most restricted numbers first:

• $7$: only $14$ works. $1$ choice.
• $6$: $12, 13, 14$ minus the one already used. $2$ choices.
• $5$: $10$ through $14$ minus two used. $3$ choices.
• $4$: $8$ through $14$ minus three used. $4$ choices.
• $3, 2, 1$: any of the remaining three, in turn. $3 \cdot 2 \cdot 1$ choices.

Total: $1 \cdot 2 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 144$.

The answer is $\boxed{\textbf{(E)}\ 144}$.

The allowed partners of $7, 6, 5, 4, \dots$ form nested sets: whatever a larger number may take, a smaller number may also take. Matching in decreasing order of the smaller member therefore removes exactly one option per step from each later set, so the count at each step is fixed and the multiplication principle applies with no casework.

• Assigning partners starting from 1, where the number of available choices depends on earlier picks, instead of starting from the most constrained number 7.
• Overlooking that $7$ must be a smaller member (since $8$ through $14$ already fill all seven "larger" slots) and allowing pairs like $(3, 7)$.