2023/24 AHSMC Part 1 Problem 8

Let $X = \{4,5,6,8,9,10,14\}$. The arrangement must start with $5,10$ and end with $6,9$, or start with $9,6$ and end with $10,5$, since $\gcd(5,x)=1$ for every $x \in X$ except $x=10$, and $\gcd(9,x)=1$ for every $x \in X$ except $x=6$. Every permutation of $4,8,14$ between the $6$ and $10$ works, since these five numbers are all even. Thus there are $2 \cdot 3! = 12$ such arrangements. The answer is (C).