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Repeated Letters in Words

Introduction

कई word-arrangement problems में repeated letters होते हैं। जब letters repeat होते हैं, तो कुछ arrangements एक जैसे दिखते हैं - इसलिए overcounting से बचने के लिए हर repeat count के factorial से divide करना ज़रूरी होता है।

यह pattern महत्वपूर्ण है क्योंकि repeats के साथ सही counting करना exams में बहुत पूछा जाता है और anagrams, license-plate type problems और arrangement puzzles में दिखाई देता है।

Pattern: Repeated Letters in Words

Pattern

अगर किसी word में n letters हों और कुछ letters repeat हों - जिनके counts p, q, r, … हों - तो distinct arrangements की संख्या:

Total = n! / (p! × q! × r! × ...)

Idea: पहले n! लें (मानकर कि सभी letters distinct हैं), फिर हर repeated-letter के factorial से divide करें क्योंकि identical letters की internal permutations नया arrangement नहीं बनातीं।

Step-by-Step Example

Question

BALLOON शब्द की letters से कितने distinct arrangements बनाए जा सकते हैं?

Solution

  1. Step 1: Total letters और repeats गिनें।

    शब्द BALLOON में n = 7 letters हैं। Letter counts: B = 1, A = 1, L = 2, O = 2, N = 1.

  2. Step 2: Formula चुनें।

    Use n! / (p! × q! × ...) जहाँ p, q repeated letters के frequencies हैं। यहाँ repeats हैं L (2 बार) और O (2 बार)।

  3. Step 3: Values substitute करें।

    Total = 7! / (2! × 2!)

  4. Step 4: Compute step-by-step.

    1. 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040.
    2. 2! = 2; इसलिए 2! × 2! = 2 × 2 = 4.
    3. Total = 5040 ÷ 4 = 1260.

  5. Final Answer:

    BALLOON के 1260 distinct arrangements बनते हैं।

  6. Quick Check:

    अगर L और O को distinct मानें, तो 7! = 5040 arrangements मिलते। फिर L के 2! और O के 2! से divide करें → 5040 ÷ (2 × 2) = 1260 ✅

Quick Variations

1. अगर तीन letters repeat हों और उनकी frequencies p, q, r हों → n! / (p! q! r!)।

2. अगर केवल किसी subset को arrange करना हो जिसमें repeats हों, तो उसी subset की frequencies गिनें और वही rule apply करें।

3. Circular arrangements में repeats होने पर पहले linear arrangements निकालें और फिर n से divide करें (अगर rotations identical हों) - symmetry और repeated blocks का ध्यान रखें।

Trick to Always Use

  • Step 1 → Total letters n और repeated letters की frequencies list करें।
  • Step 2 → पहले n! compute करें (या top r factors), फिर हर repeat count के factorial से divide करें।
  • Step 3 → Quick sanity check: answer integer होना चाहिए और n! से छोटा।

Summary

Summary

जब letters repeat हों, तो n! से शुरू करें और हर repeated-letter के factorial से divide करें:

  • Formula: n! / (p! × q! × ...).
  • सबसे पहले letter counts लिखें - इससे गलती नहीं होती।
  • Quick check: result integer हो और n! से कम हो; एक छोटा example test करके भी verify कर सकते हैं।

Practice

(1/5)
1. How many distinct arrangements can be made from the letters of the word 'MISS'?
easy
A. 12
B. 24
C. 6
D. 4

Solution

  1. Step 1: Count letters and repeats.

    The word MISS has n = 4 letters with S repeated 2 times (M=1, I=1, S=2).

  2. Step 2: Use the repeated-letter formula.

    Total arrangements = n! / (p! × q! × ...) → here 4! / 2!.

  3. Step 3: Compute.

    4! = 24; 2! = 2; arrangements = 24 ÷ 2 = 12.

  4. Final Answer:

    12 distinct arrangements → Option A.

  5. Quick Check:

    Result is integer and less than 4!; swapping the two S's does not produce a new arrangement, so 24 ÷ 2 = 12 ✅
Hint: Start with n!, then divide by factorials of repeated-letter counts.
Common Mistakes: Forgetting to divide by the factorial of repeated letters (counting S-swaps as distinct).
2. How many distinct arrangements can be formed from the letters of the word 'LEVEL'?
easy
A. 20
B. 30
C. 60
D. 24

Solution

  1. Step 1: Count letters and repeats.

    The word LEVEL has n = 5 letters with L repeated 2 times and E repeated 2 times (L=2, E=2, V=1).

  2. Step 2: Apply the formula.

    Total = 5! / (2! × 2!).

  3. Step 3: Compute.

    5! = 120; 2! × 2! = 2 × 2 = 4; arrangements = 120 ÷ 4 = 30.

  4. Final Answer:

    30 distinct arrangements → Option B.

  5. Quick Check:

    120 (if all distinct) reduced by factor 4 due to two pairs of identical letters → 30 ✅
Hint: Divide n! by the product of factorials of each letter's repeat count.
Common Mistakes: Treating each repeated letter as different and not dividing by repeat factorials.
3. How many distinct arrangements can be formed from the letters of the word 'BANANA'?
easy
A. 90
B. 120
C. 60
D. 30

Solution

  1. Step 1: Count letters and repeats.

    The word BANANA has n = 6 letters with A repeated 3 times and N repeated 2 times (A=3, N=2, B=1).

  2. Step 2: Apply the formula.

    Total = 6! / (3! × 2!).

  3. Step 3: Compute.

    6! = 720; 3! × 2! = 6 × 2 = 12; arrangements = 720 ÷ 12 = 60.

  4. Final Answer:

    60 distinct arrangements → Option C.

  5. Quick Check:

    720 divided by 12 (internal swaps from A and N) gives an integer 60 ✅
Hint: List repeat frequencies quickly: divide n! by product of those factorials.
Common Mistakes: Missing one of the repeat counts (e.g., forgetting N repeats twice).
4. How many distinct arrangements can be formed from the letters of the word 'SUCCESS'?
medium
A. 360
B. 840
C. 720
D. 420

Solution

  1. Step 1: Count letters and repeats.

    The word SUCCESS has n = 7 letters with S repeated 3 times, C repeated 2 times, and U=1, E=1 (S=3, C=2, U=1, E=1).

  2. Step 2: Apply the formula.

    Total = 7! / (3! × 2!).

  3. Step 3: Compute.

    7! = 5040; 3! × 2! = 6 × 2 = 12; arrangements = 5040 ÷ 12 = 420.

  4. Final Answer:

    420 distinct arrangements → Option D.

  5. Quick Check:

    5040 reduced by factor 12 (for internal swaps of S and C) → 420 ✅
Hint: Always count each repeated-letter frequency before dividing into n!.
Common Mistakes: Using incorrect repeat counts (e.g., treating S as twice instead of thrice).
5. How many distinct arrangements can be formed from the letters of the word 'BOOKKEEPER'?
medium
A. 1,51,200
B. 3,02,400
C. 75,600
D. 15,120

Solution

  1. Step 1: Count letters and repeats.

    The word BOOKKEEPER has n = 10 letters with counts: O=2, K=2, E=3, B=1, P=1, R=1.

  2. Step 2: Apply the formula.

    Total = 10! / (2! × 2! × 3!).

  3. Step 3: Compute.

    10! = 36,28,800; denominator = 2 × 2 × 6 = 24; arrangements = 36,28,800 ÷ 24 = 1,51,200.

  4. Final Answer:

    1,51,200 distinct arrangements → Option A.

  5. Quick Check:

    Result is integer and ≤ 10!; dividing by repeat-factor 24 reduces overcounting correctly ✅
Hint: Compute denominator as product of repeat-factorials (2!, 3!, etc.) then divide into n!.
Common Mistakes: Forgetting to include one repeated-letter factorial (e.g., missing E's 3!).

Mock Test

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