Question

Total number of $n-digit$ numbers (where, $n>1)$, having the property that no two consecutive digits are same, is

• A. $8^n$
• B. $9^n$
• C. $9\cdot {10}^{n-1}$
• D. None of these

Answer 1

The correct option is C $9\cdot {10}^{n-1}$

$\square$ $\square$ ... $\square$

$x_1$ $x_2$ $x_n$

The digit $x_1$ can be selected in $9$ ways as $0$ cannot be selected.
The digit $x_2$ can be selected in $9$ ways as $0$ can be selected but digit in positive $x_1$ cannot be selected.
Similarly, all the remaining digits can also be selected in $9$ ways each.
Thus, total number of such numbers $=9^n$.