91Ó°ÊÓ

We have to form $n$-digit numbers where each digit is one of the $10$integers $0,1,...,9$and no two consecutive digits are equal.

Total no. of digits $=10$

So, the first place can be filled in $=10!\text{ways}$$=10$

Each of the remaining $n-1$place can be filled in $=9!\text{ways}=9$as two consecutive digits cannot be equal.

Therefore, the total numbers for which no two consecutive digits are equal are$10×9^{n-1}$.

We have to form $n$-digit numbers where each digit is one of the $10$integers $0,1,...,9$and $0$ appears as a digit a total of $i$times, where $i=0,...,n$.

No. of choices of $i$places to put $0$is $=\left(\begin{array}{c}n\\ i\end{array}\right)$

The remaining $n-i$position can be filled by any of the $9$digits, $1,2,........,9$.

Therefore, the total numbers for which $0$ appears as a digit a total of $i$ times,$i=0,...,n$$=\left(\begin{array}{c}n\\ i\end{array}\right)9^{n-i}$.