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Repeated Letters in Words

Introduction

Many word-arrangement problems include repeated letters. When letters repeat, some arrangements look identical - we must divide by the factorial of each repeat count to avoid overcounting.

This pattern is important because counting correctly with repeats is a common exam topic and appears in anagrams, license-plate style problems, and arrangement puzzles.

Pattern: Repeated Letters in Words

Pattern

If a word has n letters where certain letters repeat - counts p, q, r, … - the number of distinct arrangements is:

Total = n! / (p! × q! × r! × ...)

Idea: Start with n! (arranging all letters as if distinct) then divide by the factorial of each repeated-letter frequency because each group of identical letters can be permuted internally without creating a new distinct arrangement.

Step-by-Step Example

Question

How many distinct arrangements can be formed from the letters of the word BALLOON?

Solution

  1. Step 1: Count total letters and repeats.

    The word BALLOON has n = 7 letters. Letter counts: B = 1, A = 1, L = 2, O = 2, N = 1.

  2. Step 2: Choose the formula.

    Use n! / (p! × q! × ...) where p, q are repeat frequencies. Here repeats are L (2 times) and O (2 times).

  3. Step 3: Substitute values.

    Total = 7! / (2! × 2!)

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

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

  5. Final Answer:

    There are 1260 distinct arrangements of BALLOON.

  6. Quick Check:

    If L and O were treated as distinct letters we'd get 7! = 5040 arrangements; dividing by 2! for L and 2! for O removes duplicate internal swaps → 5040 ÷ (2 × 2) = 1260 ✅

Quick Variations

1. If three letters repeat with frequencies p, q, r, use n! / (p! q! r!).

2. If you arrange only a subset containing repeats, count frequencies in that subset and apply the same division rule.

3. For circular arrangements with repeats, compute linear arrangements with repeats then divide by n (if rotations identical) - be careful with symmetry and repeated-block effects.

Trick to Always Use

  • Step 1 → Count total letters n and list frequencies of each repeated letter.
  • Step 2 → Compute n! first (or the top r factors if simplifying), then divide by the factorial of each repeat count.
  • Step 3 → Do a quick sanity check: result must be an integer and ≤ n!.

Summary

Summary

When letters repeat, start from n! and divide by factorials of each repeated-letter frequency:

  • Formula: n! / (p! × q! × ...).
  • Always list letter counts first - that prevents mistakes.
  • Quick check: result is an integer and less than or equal to n!; test with a small example to verify logic.

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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