Number of palindromes of $n$ digits that contains at least two zeros

Question

If $n$ is an odd number, determine the number of numerical palindromes of $n$ digits that contains at least two zeros.

Then I considered:

$(a,b,a) \rightarrow 10$ possibilities

$(a,b,c,b,a) \rightarrow 200$ possibilities

$(a,b,c,d,c,b,a) \rightarrow 3000$ possibilities

Then for $n = 2k+1$ there are $k10^{k}$ possibilities. Is this correct?

Answer 1

For $n=2k+1$:

Total number of palindromes: $10^{k+1}$

Number of palindromes without a zero: $9^{k+1}$

Number of palindromes with exactly one zero: $9^{k}$

Number of palindromes with at least $2$ zeros: $10^{k+1}-9^{k+1}-9^{k}$