2022 AMC 10A Problem 14

The numbers from $8$ through $14$ cannot be paired with one another, so they must be paired with the numbers from $1$ through $7.$ In particular, $7$ must be paired with $14,$ since no other available number is at least twice $7.$ Now let’s look at what the other numbers can pair with. $8$ and $9$ can pair with any number $1-4.$ $10$ and $11$ can pair with any number $1-5,$ and $12$ and $13$ can pair with any number $1-6.$ $8$ can pair with $4$ numbers, but then $9$ only has $3$ options since $8$ took one. $10$ then has $3$ options, since $2$ choices are taken, but it has one more to choose from $(5).$ $11$ then has $2$ options, $12$ has $2$ options, and $13$ only has $1.$ Multiplying these together yields $$4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 144.$$ Thus, E is the correct answer.