2022 AMC 10A Problem 14

Step 1: which numbers can be the smaller one in a pair? If $x$ is the smaller number, its partner is at least $2x$, and the partner is at most $14$. So $2x \le 14$, which means $x \le 7$.

So $8, 9, \ldots, 14$ are always the larger number of their pair. That is $7$ numbers, and there are exactly $7$ pairs, each with one larger number. So the larger numbers are exactly $8$ through $14$, and the smaller numbers are exactly $1$ through $7$. Every pair has one number from each group.

Step 2: list the allowed partners.

Step 3: match the pickiest number first. Each list contains the one above it, so every partner already taken is always in the current list, and we can simply subtract.

• $7$: $1$ choice ($14$).
• $6$: $3$ allowed, $1$ already taken: $2$ choices.
• $5$: $5$ allowed, $2$ taken: $3$ choices.
• $4$: $7$ allowed, $3$ taken: $4$ choices.
• $3$, $2$, $1$: the $3$ partners left can go to them in $3 \times 2 \times 1$ ways.

Step 4: multiply. $1 \times 2 \times 3 \times 4 \times 3 \times 2 \times 1 = 144$.

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

When items have different numbers of options, place the most restricted one first. Here the lists of allowed partners sit inside each other, so going from $7$ down to $1$ each step loses exactly one option per earlier pick. Every count is then a fixed number, and the multiplication principle works with no cases.

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 $\boxed{\textbf{(E)}\ 144}$.