The number of 'n' digit numbers such that no two consecutive digits are same is

Options

A. `n!`
B. `9!`
C. `9^(n)`
D. `n^(9)`

Answer

To find the number of 'n' digit numbers such that no two consecutive digits are the same, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to form 'n' digit numbers where no two consecutive digits are the same. The digits can range from 0 to 9.

2. **Identifying the First Digit**:
The first digit of an 'n' digit number cannot be 0 (as it would not be a valid 'n' digit number). Therefore, we have 9 options for the first digit (1 through 9).

3. **Identifying Subsequent Digits**:
For each subsequent digit (from the second digit to the nth digit), we can use any digit from 0 to 9, but it cannot be the same as the digit immediately before it. This means for each of these digits, we have 9 options (since we can choose any digit except the one that was just used).

4. **Calculating the Total Combinations**:
- The first digit has 9 choices.
- The second digit has 9 choices (it can be any digit except the first).
- The third digit also has 9 choices (it can be any digit except the second), and so on.

Therefore, the total number of 'n' digit numbers can be calculated as:
\[ \text{Total combinations} = 9 \times 9 \times 9 \times \ldots \text{(n times)} = 9^n \]

5. **Final Result**:
Thus, the number of 'n' digit numbers such that no two consecutive digits are the same is given by:
\[ \boxed{9^n} \]