2022 AMC 12A Problem 10

Any number $8$ or larger cannot be a smaller element (its double exceeds $14$), so $8$–$14$ are all larger elements and $1$–$7$ are all smaller elements. Match each smaller $s$ to a larger $g\ge2s.$ Processing from the most restrictive: $s=7$ forces $g=14$ ($1$ way); then $s=6$ has $\{12,13\}$ left ($2$); $s=5$ has $3;$ $s=4$ has $4;$ $s=3$ has $3;$ $s=2$ has $2;$ $s=1$ has $1.$ The number of matchings is $1\cdot2\cdot3\cdot4\cdot3\cdot2\cdot1=144.$ Thus, the correct answer is E.