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Repetition Allowed (Permutation with Repetition)

Introduction

Repetition Allowed pattern (जिसे permutation with repetition भी कहा जाता है) तब लागू होता है जब हर position को same item से कई बार भरा जा सकता है। यह passwords, PINs, codes जैसी situations में बहुत common है जहाँ choices repeat हो सकती हैं।

यह pattern महत्वपूर्ण है क्योंकि repetition allowed होने पर counting आसान हो जाती है - factorials की जगह exponentiation (n^r) का उपयोग करते हैं।

Pattern: Repetition Allowed (Permutation with Repetition)

Pattern

जब आपके पास हर position के लिए n choices हों और positions की संख्या r हो, तथा repetition allowed हो, तो total arrangements = nr।

Formula: Total outcomes = nr (क्योंकि r में से हर position के पास independently n choices होती हैं।)

Step-by-Step Example

Question

A, B, C, D letters का उपयोग करके repetition allowed हो, तो कितने 3-letter codes बनाए जा सकते हैं?

Solution

  1. Step 1: दिए गए values पहचानें।

    Letters = n = 4 (A, B, C, D). Code length = r = 3. Repetition allowed है।

  2. Step 2: Repetition rule लागू करें।

    हर position पर 4 letters में से कोई भी आ सकता है, इसलिए formula nr use करें।

  3. Step 3: Substitute करें और compute करें।

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

  4. Final Answer:

    Repetition allowed होने पर कुल 64 अलग-अलग 3-letter codes बनते हैं।

  5. Quick Check:

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

Quick Variations

1. Digit codes: जैसे 0-9 से 4-digit PIN → 104 = 10,000।

2. Mixed positions: letters + digits (जैसे 2 letters और 3 digits) → parts multiply करें: 26² × 10³।

3. अगर repetition allowed नहीं है: permutations (nPr) use करें - अलग formula।

Trick to Always Use

  • Step 1: Check करें कि repetition allowed है या नहीं। अगर YES → nr।
  • Step 2: Mixed structures को independent blocks में तोड़ें और multiply करें (जैसे letters block × digits block)।

Summary

Summary

जब हर position पर choices repeat हो सकती हों:

  • Use nr जहाँ n = प्रति position choices, r = positions की संख्या।
  • Passwords, PINs, codes और repeated selections में लागू।
  • Mixed formats में हर block अलग compute करें और results multiply करें।

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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