From the digits 8, 5, 2, 1, 7 and 6, how many different two digit numbers can be formed which contain the digit 8 (in either the tens place or the units place), if no digit is repeated within a number?

Introduction / Context:
This is a basic permutations and combinations problem involving the formation of two digit numbers using a given set of digits. The numbers must contain the digit 8 and no digit may be repeated within a number. Such problems test understanding of positional choices in number formation and counting valid combinations systematically.

Given Data / Assumptions:

• Available digits are: 8, 5, 2, 1, 7 and 6.
• We need to form two digit numbers.
• Each required number must contain the digit 8.
• No digit can be repeated in the same number.
• Both the tens place and the units place are allowed for the digit 8.

Concept / Approach:
To form a two digit number with no repetition and with 8 included, we must choose a partner digit for 8 from the remaining digits and then decide in which position 8 will appear. The remaining digits available to pair with 8 are 5, 2, 1, 7 and 6. For each choice of partner digit, there are two possible arrangements: 8 in the tens place or 8 in the units place. The total number of valid two digit numbers is therefore the number of choices of the partner digit multiplied by the number of possible positions for 8.

Step-by-Step Solution:
Step 1: Count the digits that can pair with 8. These are 5, 2, 1, 7 and 6, which gives 5 possible partner digits. Step 2: For each partner digit, form two numbers: one with 8 in the tens place and the partner in the units place, and one with the partner in the tens place and 8 in the units place. Step 3: For example, with partner digit 5, possible numbers are 85 and 58. With partner digit 2, numbers are 82 and 28, and similarly for 1, 7 and 6. Step 4: Thus, for each of the 5 partner digits, there are 2 distinct two digit numbers. Step 5: Total required numbers = 5 * 2 = 10.

Verification / Alternative check:
We can list all numbers explicitly: 85, 58, 82, 28, 81, 18, 87, 78, 86, 68. All of these contain the digit 8 exactly once and no digit is repeated within any number. There are no other two digit numbers satisfying the conditions, and counting them confirms that there are exactly 10 such numbers.

Why Other Options Are Wrong:
6 and 9 underestimate the total count because they do not account for both positions of 8 for every partner digit. 11 would overcount and would require an additional arrangement that is not possible under the non repetition restriction.

Common Pitfalls:
Learners sometimes forget that 8 can appear in both the tens and units positions and count only one arrangement per partner digit. Others may mistakenly include numbers like 88, which are not allowed because digits cannot be repeated within the same number. Being systematic about partner selection and position choices helps keep the count accurate.

Final Answer:
There are 10 different two digit numbers containing the digit 8.