Problem (Number Theory):
Let $T=11$. In ARMLvania, license plates use only the digits 1-9, and each license plate contains exactly $T-3$ digits. On each plate, all digits are distinct, and for all $k \leq T-3$, the $k^{\text {th }}$ digit is at least $k$. Compute the number of valid ARMLvanian license plates.

Solution:
There are 9 valid one-digit plates. For a two-digit plate to be valid, it has to be of the form $\underline{A} \underline{B}$, where $B \in\{2, \ldots, 9\}$, and either $A \in\{2, \ldots, 9\}$ with $A \neq B$ or $A=1$. So there are 8 ways to choose $B$ and $8-1+1=8$ ways to choose $A$, for a total of $8 \cdot 8=64$ plates. In general, moving from the last digit to the first, if there are $k$ ways to choose digit $n$, then there are $k-1$ ways to choose digit $n-1$ from the same set of possibilities as digit $n$ had, plus one additional way, for a total of $k-1+1=k$ choices for digit $n-1$. So if a license plate has $d$ digits, there are $10-d$ choices for the last digit and for each digit before it, yielding $(10-d)^{d}$ possible $d$-digit plates. Using $d=T-3=8$, there are $2^{8}=\mathbf{2 5 6}$ plates.

Final answer: 256