jmc

There are $9$ choices for the first number, since it cannot be $0$, there are only $9$ choices left for the second number since it must differ from the first, $8$ choices for the third number, since it must differ from the first two, and $7$ choices for the fourth number, since it must differ from all three. This means there are $9\times 9\times 8\times 7=\boxed{\text{(B) }4536}$ integers between $1000$ and $9999$ with four distinct digits. If you're too lazy to do $9\times 9\times 8\times 7=4536$, we can notice $9\times 9=81$ and $8\times 7=56$ The unit digit must be 6. Therefore the answer is the only answer with a a unit digit of 6. $\boxed{4536}$)