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Question

(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 $0$). Thus,there are $9$ c

Vedclass · Accepted Answer

$9 	imes 9^{n-1}$

Vedclass · Answer

(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 $0$). Thus,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. Thus,there are $9$ choices for the second digit.For the third digit,it can be any digit from $0$ to $9$ except the one used for the second digit. Thus,there are $9$ choices for the third digit.Continuing this pattern,for each subsequent position up to the $n$-th digit,there are always $9$ choices available because each digit only needs to be different from the one immediately preceding it.Therefore,the total number of $n$-digit numbers is $9 	imes 9 	imes 9 	imes \dots 	imes 9$ ($n$ times),which equals $9 	imes 9^{n-1}$.

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

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