Q. Assuming that no two consecutive digits are same. The number of $n$ digit numbers is
A) $n!$
B) $9!$
C) $9^{n}$
D) $n^{9}$
Explanation
(c)
Digit at the extreme left can be chosen by 9 ways as zero cannot be the first digit. Now for the second digit it can be done in 9 ways as consecutive digits are not same. And this is same for next digits. Hence, number of ways are
$9 \times 9 \times 9 \times \ldots.. \times n$ times$= 9^{n}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.