2006 AMC 10A Problem 21

"At least one" suggests counting the opposite: four-digit numbers that contain no $2$ and no $3$ at all.

Total four-digit numbers: the leading digit is $1$ through $9$ and each other digit is $0$ through $9$, so $9 \cdot 10 \cdot 10 \cdot 10 = 9000$.

Numbers avoiding $2$ and $3$: the leading digit comes from $\{1, 4, 5, 6, 7, 8, 9\}$, $7$ options; each of the other three digits comes from the eight digits other than $2$ and $3$. That gives $7 \cdot 8 \cdot 8 \cdot 8 = 7 \cdot 512 = 3584$.

Subtract: $9000 - 3584 = 5416$.

The answer is $\boxed{\textbf{(E)}\ 5416}$.

A condition of the form "at least one digit is special" splits into many overlapping cases if attacked directly, but its complement, "no digit is special," is a single product of independent choices. The only care needed is the leading digit, which loses the option $0$ in the total and loses $0$, $2$, $3$ in the complement.

Rough check: each digit is a $2$ or $3$ with probability about $\tfrac15$, so a random four-digit number avoids both with probability about $\left(\tfrac45\right)^4 \approx 0.41$; roughly $59\%$ of $9000$, about $5300$, should qualify, and only choice (E) is close.

• Allowing $0$ as a leading digit in the complement ($8^4 = 4096$, choice (B)) or subtracting from $10^4$ instead of $9000$.
• Trying direct casework on how many $2$s and $3$s appear and losing track of the overlapping cases.