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

Options

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

Explanation

To solve the problem of finding the number of three-digit numbers having only two consecutive digits identical, we can break it down into two cases based on the positions of the identical digits.

### Step-by-Step Solution:

**Step 1: Define the Cases**
We will consider two cases:
- Case 1: The first two digits are identical (A, A, B).
- Case 2: The last two digits are identical (A, B, B).

**Step 2: Case 1 - Digits A, A, B**
- Here, the first digit (A) cannot be 0 because it is a three-digit number. Therefore, A can take any value from 1 to 9 (9 options).
- The second digit (A) is the same as the first digit.
- The third digit (B) can be any digit from 0 to 9, but it must be different from A. Therefore, B has 9 options (0-9 excluding A).
- The total combinations for this case can be calculated as:
\[ \text{Total for Case 1} = 9 \text{ (choices for A)} \times 9 \text{ (choices for B)} = 81 \]

**Step 3: Case 2 - Digits A, B, B**
- In this case, the first digit (A) again cannot be 0, so A can take any value from 1 to 9 (9 options).
- The second digit (B) can be any digit from 0 to 9, but it must be different from A. Therefore, B has 9 options (0-9 excluding A).
- The total combinations for this case can be calculated as:
\[ \text{Total for Case 2} = 9 \text{ (choices for A)} \times 9 \text{ (choices for B)} = 81 \]

**Step 4: Combine the Results**
- Now, we add the results from both cases to find the total number of three-digit numbers with only two consecutive digits identical:
\[ \text{Total} = \text{Total for Case 1} + \text{Total for Case 2} = 81 + 81 = 162 \]

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

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