The number of three-digit numbers having only two consecutive digits identical is

Options

A. `153`
B. `162`
C. `180`
D. `161`

Explanation

To find the number of three-digit numbers having only two consecutive digits identical, we will break down the problem into two cases based on the position of the identical digits.

### Step-by-Step Solution:

**Case 1: The first two digits are identical (A, A, B)**

1. **Choose A**: The first digit (A) cannot be 0 (since it's a three-digit number). Therefore, A can be any digit from 1 to 9. This gives us 9 options for A.

2. **Choose B**: The third digit (B) can be any digit from 0 to 9, but it must be different from A. Since A has already taken one digit, we have 9 remaining options for B.

3. **Total combinations for Case 1**:
\[ \text{Total for Case 1} = \text{options for A} \times \text{options for B} = 9 \times 9 = 81 \]

**Case 2: The last two digits are identical (A, B, B)**

1. **Choose A**: The first digit (A) again cannot be 0. So, A can be any digit from 1 to 9, giving us 9 options for A.

2. **Choose B**: The last digit (B) can be any digit from 0 to 9, but it must be different from A. Thus, we again have 9 remaining options for B.

3. **Total combinations for Case 2**:
\[ \text{Total for Case 2} = \text{options for A} \times \text{options for B} = 9 \times 9 = 81 \]

**Final Calculation**:
To find the total number of three-digit numbers with only two consecutive digits identical, we add the totals from both cases:
\[ \text{Total} = \text{Total for Case 1} + \text{Total for Case 2} = 81 + 81 = 162 \]

### Conclusion:
The total number of three-digit numbers having only two consecutive digits identical is **162**.