A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected...

1. Translate the problem requirements: We need to count all possible stock codes that can be formed using 1, 2, or 3 letters from the alphabet, where letters can be repeated and order matters (AB is different from BA).
2. Count one-letter codes: Determine how many different codes can be made using exactly one letter.
3. Count two-letter codes: Determine how many different codes can be made using exactly two letters, considering that letters can repeat and order matters.
4. Count three-letter codes: Determine how many different codes can be made using exactly three letters, applying the same principles.
5. Sum all possibilities: Add up the counts from all three categories to get the total number of unique stock codes possible.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're looking for in simple terms. We have a stock exchange that creates codes for stocks using letters from the alphabet. Think of it like creating license plates, but with letters only.

Here are the key rules:

• Each code can have 1, 2, or 3 letters
• We can use any of the 26 letters in the alphabet
• We can repeat letters (so "AA" is allowed)
• Order matters (so "AB" is different from "BA")

We need to count ALL possible codes that follow these rules.

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

2. Count one-letter codes

For one-letter codes, this is straightforward. We can pick any letter from A to Z.

Since there are 26 letters in the alphabet, and we can use any one of them, we have:
26 one-letter codes

Examples: A, B, C, ..., Z

3. Count two-letter codes

For two-letter codes, let's think about this step by step:

• For the first position: We can choose any of the 26 letters
• For the second position: We can also choose any of the 26 letters (remember, repetition is allowed)

Since we make these choices independently, and each choice for the first letter can be combined with each choice for the second letter:
\(26 \times 26 = 676\) two-letter codes

Examples: AA, AB, AC, ..., BA, BB, BC, ..., ZZ

This makes sense because we're essentially filling two positions, and each position has 26 possibilities.

4. Count three-letter codes

For three-letter codes, we apply the same logic:

• First position: Any of the 26 letters
• Second position: Any of the 26 letters
• Third position: Any of the 26 letters

Since all three choices are independent:
\(26 \times 26 \times 26 = 17,576\) three-letter codes

Examples: AAA, AAB, AAC, ..., ZZZ

Process Skill: APPLY CONSTRAINTS - Ensuring we account for repetition being allowed and order mattering

5. Sum all possibilities

Now we add up all the different types of codes:

• One-letter codes: 26
• Two-letter codes: 676
• Three-letter codes: 17,576

Total = \(26 + 676 + 17,576 = 18,278\)

Let's verify this calculation:
\(26 + 676 = 702\)
\(702 + 17,576 = 18,278\) ✓

Final Answer

The total number of different stocks that can be uniquely designated is 18,278.

This matches answer choice E. 18278.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding that order matters

Students often overlook the phrase "the same letters used in a different order constitute a different code." This is crucial because it means AB and BA are different codes. If students miss this, they might try to use combinations instead of permutations, significantly undercounting the possibilities. For example, for two-letter codes, they might think there are only \(26 \times 25/2 = 325\) codes instead of \(26 \times 26 = 676\).

2. Forgetting that repetition is allowed

The problem states "letters may be repeated," but students sometimes miss this detail. If they assume letters cannot be repeated, they would calculate two-letter codes as \(26 \times 25 = 650\) (instead of \(26 \times 26 = 676\)) and three-letter codes as \(26 \times 25 \times 24 = 15,600\) (instead of \(26 \times 26 \times 26 = 17,576\)). This would lead to a significantly different final answer.

3. Not recognizing this as separate counting problems

Some students might try to find a single formula for all codes instead of recognizing that one-letter, two-letter, and three-letter codes are separate, independent counting problems that need to be solved individually and then added together.

Errors while executing the approach

1. Arithmetic calculation errors

Students might make computational mistakes, especially when calculating \(26^3 = 17,576\). They might incorrectly calculate this as 17,556 or another close number, leading to a wrong final sum.

2. Addition errors when summing the totals

Even with correct individual calculations (26, 676, and 17,576), students might make errors when adding these together. For instance, they might get 18,268 instead of 18,278, or make other similar addition mistakes.

Errors while selecting the answer

No likely faltering points - the calculation directly gives a numerical result that matches one of the answer choices, with no additional interpretation or conversion needed.