jmc

For his first digit, Benjamin has 10 possible choices. For his second digit, he has 9 possible choices, since he cannot repeat any digits. For his third, fourth, and fifth digits, he has 8, 7, and 6 possible choices. Therefore, there are $10\cdot 9\cdot 8\cdot 7\cdot 6=\boxed{30,240}$ possible codes.