How many combinations of two-digit numbers having 8 can be made from the following numbers?
8, 5, 2, 1, 7,6

Options

A. 9
B. 10
C. 11
D. 12

Explanation

To find how many combinations of two-digit numbers can be formed that include the digit 8 from the given numbers (8, 5, 2, 1, 7, 6), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Digits**: The available digits are 8, 5, 2, 1, 7, and 6.

2. **Determine the Position of 8**: Since we want to form two-digit numbers that must include the digit 8, we can place 8 in either the tens place or the units place.

3. **Case 1: 8 in the Tens Place**:
- If 8 is in the tens place, the number will be of the form 8X, where X can be any of the other digits (5, 2, 1, 7, 6).
- Possible combinations:
- 85
- 82
- 81
- 87
- 86
- Total combinations for this case: **5** (85, 82, 81, 87, 86)

4. **Case 2: 8 in the Units Place**:
- If 8 is in the units place, the number will be of the form X8, where X can be any of the other digits (5, 2, 1, 7, 6).
- Possible combinations:
- 58
- 28
- 18
- 78
- 68
- Total combinations for this case: **5** (58, 28, 18, 78, 68)

5. **Combine the Results**:
- Total combinations = Combinations with 8 in the tens place + Combinations with 8 in the units place
- Total combinations = 5 (from Case 1) + 5 (from Case 2) = **10**

### Final Answer:
The total number of combinations of two-digit numbers having 8 is **10**.