Number of $n$-digit palindromes

Details depend on whether for example $0110$ counts as a $4$-digit palindrome. We will suppose it doesn't. This makes things a little harder.

If $n$ is even, say $n=2m$, the first digit can be any of $9$, then the next $m-1$ can be any of $10$, and then the rest are determined. So there are $9\cdot 10^{m-1}$ palindromes with $2m$ digits.

If $n$ is odd, say $n=2m+1$, then the same sort of reasoning yields the answer $9\cdot 10^{m}$.

An n-digit number abc...xyz can be mapped to the 2n-digit palindrome abc...xyzzyx...bca, and to the (2n-1)-digit palindrome abc...xyzyx...bca. So the number of 2n-digit palindromes and (2n-1)-digit palindromes is simply the number of n-digit numbers: $9 \times 10^{n-1}$.