All of the telephone extensions in a certain company are 4-digit numbers. How many different 4-digit telephone extensions can be...

1. Translate the problem requirements: We need to form 4-digit telephone extensions where digit 3 appears exactly twice, digit 5 appears exactly once, and digit 8 appears exactly once. This gives us the digits 3, 3, 5, 8 to arrange in different orders.
2. Identify the arrangement challenge: Since we have repeated digits (two 3's), we cannot simply count all possible arrangements as if all digits were different. We need to account for the fact that swapping the two 3's doesn't create a new arrangement.
3. Calculate total arrangements without restrictions: First find how many ways we can arrange 4 positions if all digits were different, then adjust for the identical 3's.
4. Adjust for identical digits: Divide by the number of ways the repeated digits can be arranged among themselves to avoid overcounting.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we need to create: 4-digit telephone extensions where each extension must contain exactly two 3's, exactly one 5, and exactly one 8.

So we're arranging these four digits: 3, 3, 5, 8

Think of it like having four slots: _ _ _ _

We need to place our digits 3, 3, 5, 8 into these four positions in different ways to create different telephone extensions.

Process Skill: TRANSLATE - Converting the word problem into a concrete arrangement task

2. Identify the arrangement challenge

Here's the key insight: since we have two identical 3's, we need to be careful about counting.

If all our digits were different (like 3, 4, 5, 8), then swapping any two digits would give us a genuinely different arrangement. But since our two 3's are identical, swapping them doesn't create a new telephone extension.

For example: 3358 and 3358 are the same number, even if we got there by swapping the first and second 3's.

Process Skill: INFER - Recognizing that identical digits create overcounting issues

3. Calculate total arrangements without restrictions

Let's first imagine all four digits were different and count those arrangements.

For the first position: we have 4 choices
For the second position: we have 3 remaining choices
For the third position: we have 2 remaining choices
For the fourth position: we have 1 remaining choice

Total arrangements = \(4 \times 3 \times 2 \times 1 = 24\) arrangements

This gives us 24 different ways to arrange four distinct objects.

4. Adjust for identical digits

Now we need to account for the fact that our two 3's are identical.

In our count of 24, we've been treating the two 3's as if they were different (like 3₁ and 3₂). But they're actually the same digit.

For every arrangement we counted, we counted it twice - once with 3₁ in the first position and 3₂ in the second, and once with them swapped.

Since the two 3's can be arranged among themselves in 2 ways (3₁3₂ or 3₂3₁), and these represent the same actual arrangement, we need to divide our total by 2.

Final count = \(24 \div 2 = 12\)

Process Skill: APPLY CONSTRAINTS - Adjusting our count to reflect the constraint of identical digits

4. Final Answer

We can form 12 different 4-digit telephone extensions using the digits 3, 5, and 8, where 3 appears twice and 5 and 8 each appear once.

The answer is C. 12

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constraint requirements

Students often misread the problem and think they need to use each digit (3, 5, 8) exactly once, missing that the digit 3 appears twice. This leads them to try arranging only 3 digits instead of 4, or attempting to create 3-digit numbers instead of 4-digit telephone extensions.

2. Treating this as a combination problem instead of permutation

Some students focus on "selecting" digits rather than "arranging" them into different telephone extensions. Since the order matters for telephone extensions (3358 is different from 5833), this is a permutation problem where position matters, not a combination problem.

3. Not recognizing the identical digits issue

Students may plan to simply calculate \(4! = 24\) without realizing that having two identical 3's creates an overcounting problem. They fail to account for the fact that swapping identical digits doesn't create a new arrangement.

Errors while executing the approach

1. Arithmetic errors in factorial calculations

Students may incorrectly calculate \(4!\) as something other than 24, or make errors when dividing by \(2! = 2\) to account for the identical digits.

2. Forgetting to divide by 2! for identical digits

Even if students recognize the need to use permutations with repetition, they might calculate \(4! = 24\) but forget to divide by \(2! = 2\) to account for the two identical 3's, leading to the wrong answer of 24.

3. Incorrectly applying the repetition formula

Some students might try to use the formula \(\mathrm{n!/(n_1! \times n_2! \times ...)}\) but incorrectly divide by \(3!\) instead of \(2!\), thinking they need to account for all three different digit types rather than just the repeated 3's.

Errors while selecting the answer

No likely faltering points - the calculation directly yields 12, which matches answer choice C exactly.