A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many...

1. Translate the problem requirements: We need 3-digit numbers where all digits are non-zero (1-9) and exactly two digits are the same (meaning one digit appears twice, the other appears once)
2. Identify the two possible patterns: Since exactly two digits are the same, we can have AAB pattern (first two digits same) or ABA pattern (first and third same) or ABB pattern (last two digits same)
3. Count possibilities for each pattern systematically: For each pattern, determine how many ways we can choose the repeated digit and the different digit, then account for all arrangements
4. Sum all valid arrangements: Add up the counts from all patterns to get the total number of such 3-digit integers

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we need: We want 3-digit numbers where all digits are non-zero (so we can only use digits 1, 2, 3, 4, 5, 6, 7, 8, 9) and exactly two digits are the same.

What does "exactly two digits are the same" mean? It means one digit appears twice and another digit appears once. For example, 223 works (the digit 2 appears twice, the digit 3 appears once), but 222 doesn't work (all three digits are the same) and 123 doesn't work (no digits are repeated).

Process Skill: TRANSLATE - Converting the problem language into clear mathematical requirements

2. Identify the two possible patterns

Since we have exactly two digits that are the same in a 3-digit number, let's think about where these repeated digits can be positioned:

- Pattern 1: AAB (first two positions have the same digit) - like 223, 114, 557
- Pattern 2: ABA (first and third positions have the same digit) - like 232, 141, 575
- Pattern 3: ABB (last two positions have the same digit) - like 233, 144, 577

These are the only three ways to arrange exactly two identical digits in three positions.

Process Skill: CONSIDER ALL CASES - Systematically identifying all possible arrangements

3. Count possibilities for each pattern systematically

Now let's count how many numbers we can make for each pattern:

Pattern 1: AAB (like 223)
- Choose the repeated digit A: 9 choices (digits 1-9)
- Choose the different digit B: 8 choices (any digit 1-9 except A)
- Total for Pattern 1: \(9 \times 8 = 72\)

Pattern 2: ABA (like 232)
- Choose the repeated digit A: 9 choices (digits 1-9)
- Choose the different digit B: 8 choices (any digit 1-9 except A)
- Total for Pattern 2: \(9 \times 8 = 72\)

Pattern 3: ABB (like 233)
- Choose the different digit A: 9 choices (digits 1-9)
- Choose the repeated digit B: 8 choices (any digit 1-9 except A)
- Total for Pattern 3: \(9 \times 8 = 72\)

Notice that for each pattern, we're choosing 2 different digits from our set of 9 non-zero digits, and one of those digits gets used twice while the other gets used once.

4. Sum all valid arrangements

Now we add up all the possibilities from each pattern:

Total = Pattern 1 + Pattern 2 + Pattern 3
Total = \(72 + 72 + 72 = 216\)

Let's verify with a quick example: If we pick digits 2 and 3, we can make:
- Pattern AAB: 223
- Pattern ABA: 232
- Pattern ABB: 233
- We could also use 3 as the repeated digit: 332, 323, 322

So from just two digits (2 and 3), we get 6 different 3-digit numbers, which matches our systematic counting.

The total number of 3-digit positive integers with non-zero digits where exactly two digits are the same is 216.

This matches answer choice C. 216.