How many 3-letter codes can be formed by choosing, without replacement, 3 letters from the word PEPPER?

1. Translate the problem requirements: We need to form 3-letter codes using letters from PEPPER without replacement. This means once we use a letter, we can't use that same letter again in the same code. PEPPER has 6 letters: P, E, P, P, E, R with frequencies P(3), E(2), R(1).
2. Identify distinct letter patterns: Since we have repeated letters, we need to categorize possible 3-letter codes based on which distinct letters appear and how many times each appears.
3. Count codes for each pattern type: For each pattern (like using 3 different letters, or 2 of one letter plus 1 of another), calculate how many actual codes can be formed considering the available quantities of each letter.
4. Sum all valid combinations: Add up the counts from all possible patterns to get the total number of 3-letter codes.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're working with. The word PEPPER has 6 letters total, but some letters repeat. Let me count each distinct letter:

• P appears 3 times
• E appears 2 times
• R appears 1 time

We need to form 3-letter codes "without replacement," which means once we pick a letter from our available letters, we can't use that exact same letter again. For example, if we use one P, we still have 2 P's left to choose from.

Process Skill: TRANSLATE - Converting the problem language into a clear inventory of available letters

2. Identify distinct letter patterns

Since we're picking 3 letters and some letters repeat, let's think about what types of 3-letter codes are possible. We need to consider how many times each distinct letter (P, E, R) can appear in our code:

Pattern Type 1: Three different letters (like PER)

• We'd use P, E, and R (one of each)
• This is possible since we have at least one of each letter

Pattern Type 2: Two of one letter, one of another

• Two P's and one other letter (PP + E or PP + R)
• Two E's and one other letter (EE + P or EE + R)
• We can't do two R's since we only have one R

Pattern Type 3: Three of the same letter

• Three P's (PPP) - this is possible since we have 3 P's
• We can't do three E's or three R's since we don't have enough

Process Skill: CONSIDER ALL CASES - Systematically identifying all possible patterns based on our letter inventory

3. Count codes for each pattern type

Now let's count how many actual 3-letter codes we can make for each pattern:

Pattern 1: Three different letters (P, E, R)

• We're using one P, one E, and one R
• We can arrange these 3 distinct letters in different orders
• Number of arrangements = \(3! = 3 \times 2 \times 1 = 6\) codes
• These are: PER, PRE, EPR, ERP, RPE, REP

Pattern 2a: Two P's and one E

• We're using 2 P's and 1 E
• We need to arrange PPE in different ways
• The two P's are identical, so we have arrangements of PPE: PPE, PEP, EPP
• That's 3 codes

Pattern 2b: Two P's and one R

• We're using 2 P's and 1 R
• Arranging PPR: PPR, PRP, RPP
• That's 3 codes

Pattern 2c: Two E's and one P

• We're using 2 E's and 1 P
• Arranging EEP: EEP, EPE, PEE
• That's 3 codes

Pattern 2d: Two E's and one R

• We're using 2 E's and 1 R
• Arranging EER: EER, ERE, REE
• That's 3 codes

Pattern 3: Three P's

• We're using all 3 P's
• There's only one way to arrange PPP
• That's 1 code

4. Sum all valid combinations

Let's add up all the codes we found:

• Pattern 1 (P, E, R): 6 codes
• Pattern 2a (P, P, E): 3 codes
• Pattern 2b (P, P, R): 3 codes
• Pattern 2c (E, E, P): 3 codes
• Pattern 2d (E, E, R): 3 codes
• Pattern 3 (P, P, P): 1 code

Total: \(6 + 3 + 3 + 3 + 3 + 1 = 19\) codes

Final Answer

We can form exactly 19 different 3-letter codes from the letters in PEPPER without replacement. The answer is C) 19.

Common Faltering Points

Errors while devising the approach

• Misinterpreting "without replacement": Students often confuse this with "without repetition," thinking they can only use each distinct letter (P, E, R) once. They miss that "without replacement" means once you pick one specific letter from the available pool, that particular letter is gone, but you can still use other instances of the same letter type. For example, after using one P, you still have two more P's available.
• Failing to systematically identify all possible patterns: Students may jump straight into calculations without first mapping out what types of 3-letter combinations are actually possible given the letter constraints (3 P's, 2 E's, 1 R). This leads to missing entire categories of valid codes, such as forgetting about the PPP pattern or the double-letter combinations like PPE or EER.
• Treating identical letters as distinguishable: Students might incorrectly think that the three P's in PEPPER are different from each other (like P₁, P₂, P₃), leading them to overcomplicate the counting process. They don't realize that P-P-E and P-P-E are the same code regardless of which specific P's were chosen from the original word.

Errors while executing the approach

• Incorrect arrangement counting for repeated letters: When calculating arrangements like PPE, students often use \(3! = 6\) instead of the correct formula \(3!/2! = 3\). They forget to divide by the factorial of repeated elements, leading to overcounting. For instance, they might count PPE, PEP, EPP, PPE, PEP, EPP as six different codes when there are actually only three.
• Missing or double-counting pattern subcases: Students may forget to count certain subcases (like missing EER when they remember PPE and PPR) or accidentally count the same pattern twice. The systematic nature of checking all possible two-letter combinations (PP+E, PP+R, EE+P, EE+R) is often where students lose track.
• Arithmetic errors in final summation: When adding up all the individual pattern counts \((6 + 3 + 3 + 3 + 3 + 1)\), students may make simple addition mistakes or forget to include one of the subcategories in their final total, especially given the multiple small numbers involved.

Errors while selecting the answer

No likely faltering points - the calculation directly yields a whole number that matches one of the given options, making answer selection straightforward once the computation is complete.