Assuming that no two consecutive digits are s | vedclass

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(A) For an $n$-digit number,the first digit (at the leftmost position) can be any digit from $1$ to $9$ (since it cannot be $0$). So,there are $9$ cho

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$9 	imes 9^{n-1}$

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(A) For an $n$-digit number,the first digit (at the leftmost position) can be any digit from $1$ to $9$ (since it cannot be $0$). So,there are $9$ choices for the first digit.For the second digit,it can be any digit from $0$ to $9$ except the one used for the first digit. So,there are $9$ choices.For the third digit,it can be any digit from $0$ to $9$ except the one used for the second digit. So,there are $9$ choices.Continuing this pattern,for each of the remaining $(n-1)$ positions,there are $9$ choices.Therefore,the total number of $n$-digit numbers is $9 	imes 9 	imes 9 	imes \dots 	imes 9$ ($(n-1)$ times) $= 9 	imes 9^{n-1}$.

Assuming that no two consecutive digits are same,the number of $n$-digit numbers is:

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