2025 AMC 10A Problem 12

Place the single even digit in one of $4$ positions and multiply by $4;$ the other three digits are odd

Split on whether the one prime digit is the even one ($2$) or an odd one ($3, 5,$ or $7$)

No digit is $0,$ so digits run from $1$ to $9.$ Put the single even digit in the first slot for now and multiply by $4$ at the end to place it. Split on that even digit. If it’s the prime $2,$ then the three odd digits all have to be non-prime, so each is $1$ or $9,$ giving $2^3 = 8$ ways. Otherwise the even digit is $4, 6,$ or $8$ ($3$ choices), and exactly one of the odd digits is prime, worth $3, 5,$ or $7$ ($3$ choices) in one of the $3$ odd positions, while the other two odds come from $\{1, 9\}$ ($2^2$ ways). That’s $3 \cdot 3 \cdot 3 \cdot 4 = 108.$ Altogether, $4(8 + 108) = 464.$ Therefore, the answer is D.