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Repetition Allowed (Permutation with Repetition)

Introduction

The Repetition Allowed pattern (also called permutation with repetition) applies when each position can be filled by the same item multiple times. This is common in passwords, PINs, codes, and situations where choices can repeat.

This pattern is important because it simplifies counting when repetition is allowed - instead of factorials, we use exponentiation: n^r.

Pattern: Repetition Allowed (Permutation with Repetition)

Pattern

When you have n choices for each of r positions and repetition is allowed, the total number of arrangements is nr.

Formula: Total outcomes = nr (Each of the r positions independently has n choices.)

Step-by-Step Example

Question

How many 3-letter codes can be formed using the letters A, B, C, D if repetition is allowed?

Solution

  1. Step 1: Identify what is given.

    There are n = 4 letters (A, B, C, D). The code length is r = 3. Repetition is allowed.

  2. Step 2: Apply the repetition rule.

    Since each position can be any of the 4 letters independently, use the formula nr.

  3. Step 3: Substitute and compute.

    Total codes = 43 = 4 × 4 × 4 = 64.

  4. Final Answer:

    There are 64 different 3-letter codes when repetition is allowed.

  5. Quick Check:

    Think: For the 1st letter 4 options, 2nd letter 4 options, 3rd letter 4 options → 4 × 4 × 4 = 64 ✅

Quick Variations

1. Codes with digits: e.g., 4-digit PIN from 0-9 → 104 = 10,000.

2. Mixed positions: letters + digits (e.g., 2 letters then 3 digits) → multiply parts: 26² × 10³.

3. Repetition not allowed: use permutations (nPr) instead - distinct formula.

Trick to Always Use

  • Step 1: Check if repetition is allowed. If YES → use nr.
  • Step 2: Break mixed structures into independent blocks and multiply (e.g., letters block × digits block).

Summary

Summary

When each position can repeat choices:

  • Use nr where n = choices per position, r = number of positions.
  • Apply to passwords, PINs, codes, and independent repeated selections.
  • For mixed formats, compute each block separately and multiply results.

Practice

(1/5)
1. How many 3-digit PINs can be formed using digits 0-9 if repetition is allowed?
easy
A. 1000
B. 900
C. 720
D. 10000

Solution

  1. Step 1: Identify what is given.

    Digits available = 10 (0-9). Number of positions r = 3. Repetition allowed.

  2. Step 2: Apply the repetition rule.

    Total PINs = nr = 103.

  3. Step 3: Substitute and compute.

    103 = 1000.

  4. Final Answer:

    There are 1000 possible 3-digit PINs → Option A.

  5. Quick Check:

    Each of 3 positions has 10 choices → 10 × 10 × 10 = 1000 ✅
Hint: When repetition allowed, use n^r (choices^positions).
Common Mistakes: Using permutations (nPr) or forgetting that 0 is allowed in the first position.
2. How many 2-letter codes can be formed using the 26 English letters if repetition is allowed?
easy
A. 325
B. 676
C. 650
D. 702

Solution

  1. Step 1: Identify what is given.

    Letters available = 26. Positions r = 2. Repetition allowed.

  2. Step 2: Apply the rule for repetition.

    Total codes = 262.

  3. Step 3: Compute.

    26 × 26 = 676.

  4. Final Answer:

    There are 676 2-letter codes → Option B.

  5. Quick Check:

    Each position has 26 choices → 26² = 676 ✅
Hint: Compute block-wise: letters block = 26^positions.
Common Mistakes: Multiplying by 25 (no repetition) when repetition is allowed.
3. How many distinct 4-bit binary strings exist (bits can repeat)?
easy
A. 8
B. 12
C. 16
D. 24

Solution

  1. Step 1: Identify what is given.

    Choices per position (bit) = 2 (0 or 1). Positions r = 4. Repetition allowed.

  2. Step 2: Apply n^r rule.

    Total strings = 24.

  3. Step 3: Compute.

    2 × 2 × 2 × 2 = 16.

  4. Final Answer:

    There are 16 distinct 4-bit strings → Option C.

  5. Quick Check:

    Each of 4 positions has 2 choices → 2⁴ = 16 ✅
Hint: Use 2^r for binary-length strings.
Common Mistakes: Confusing with permutations without repetition (nPr).
4. A vehicle plate has 2 letters followed by 2 digits. If repetition is allowed for both letters and digits, how many plates are possible? (26 letters, 10 digits)
medium
A. 6,760
B. 67,60,000
C. 6,76,000
D. 67,600

Solution

  1. Step 1: Identify what is given.

    Letters positions = 2 (each 26 choices); digit positions = 2 (each 10 choices). Repetition allowed.

  2. Step 2: Compute blocks separately and multiply.

    Letter block = 26² = 676. Digit block = 10² = 100. Total = 676 × 100.

  3. Step 3: Compute final value.

    676 × 100 = 67,600.

  4. Final Answer:

    There are 67,600 possible plates → Option D.

  5. Quick Check:

    26² × 10² = 676 × 100 = 67,600 ✅
Hint: Multiply independent blocks: letters^positions × digits^positions.
Common Mistakes: Misplacing zeros or using 26 × 10 instead of exponent rules.
5. How many 3-character codes can be formed using the letters {A, B, C, D, E} if repetition is allowed?
medium
A. 125
B. 60
C. 75
D. 100

Solution

  1. Step 1: Identify what is given.

    Choices per position n = 5 (A-E). Positions r = 3. Repetition allowed.

  2. Step 2: Apply the repetition formula.

    Total = 53.

  3. Step 3: Compute.

    5 × 5 × 5 = 125.

  4. Final Answer:

    There are 125 possible 3-character codes → Option A.

  5. Quick Check:

    Each of 3 positions has 5 choices → 5³ = 125 ✅
Hint: Compute n^r directly for small n and r.
Common Mistakes: Using permutations (nPr) or forgetting to allow repetition.

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