The first digit can not be $0$ and so for the first digit we have 9 options (from $1$ to $9$). Now we have to choose the next digit in such a way that it is not the same as the previous one. Thereofore on choosing the next digit we can have 9 options as well, (because we can choose any number from $0$ to $9$ except the previous one). Therefore, each of the remaining 19 digits can only have 9 options. Therefore there are $9\times 19^9$ numbers are available of the given type.