2006 AMC 10A Problem 21

There are $9000$ four-digit integers. For those avoiding $2$ and $3,$ the leading digit is one of $\{1, 4, 5, 6, 7, 8, 9\}$ ($7$ choices) and each remaining digit is one of $\{0, 1, 4, 5, 6, 7, 8, 9\}$ ($8$ choices): $7 \cdot 8^3 = 3584.$ So $9000 - 3584 = 5416$ have at least one $2$ or $3.$ Thus, the correct answer is E.