Question

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

• A. $n!$
• B. $9!$
• C. $9^n$
• D. $n^9$

Answer 1

The correct option is C

$9^n$

Explanation for the correct answer:

Finding the number of $n$ digits.

There are totally $10$ digits, numbered from $0to9$.

In first position we cannot place $0$, since no numbers can start with zero. So, it has $9$ possibilities except zero.

It is given, no two consecutive digits are same.

In second position, including $0$ and excluding the previous digit, there are $9$ possibilities.

Similarly for the third, fourth,…. upto $n$ positions, there are $9$ possibilities.

Thus we can say, $9\times 9\times 9.......\times 9$ upto $n$ positions.

Therefore we have $9^n$ possibilities for $n$ digits, where no two consecutive digits are same.

Hence, option (C) ${\mathbf{9}}^{\mathbf{n}}$ is the correct answer.