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CIE A-Level Maths Study Notes

4.2.2 Arrangements with Repetition

This section focuses on understanding and solving problems where some elements are repeated, a common scenario in mathematical and practical situations.

Understanding Permutations with Repetition

Permutations deal with the arrangement of objects in a sequence, and when some of these objects are identical, we encounter permutations with repetition.

  • Definition: Permutations with repetition are arrangements of 'n' objects where some items are identical.
  • Key Concept: The presence of identical items in permutations reduces the number of unique arrangements.

Formula for Permutations with Repetition

  • Formula: Arrangements=n!n1!×n2!××nk!\text{Arrangements} = \frac{n!}{n_1! \times n_2! \times \ldots \times n_k!}
    • nn: Total objects.
    • n1,n2,,nkn_1, n_2, \ldots, n_k: Repeats of each object.

Example 1: Word Arrangements

Problem: Calculate the distinct arrangements of the word 'NEEDLESS'.


  • Total letters (n): 8 (N, E, E, D, L, E, S, S)
  • Repeated letters: E (3 times), S (2 times)


  • Total permutations without repetition:
    • 8!=40,3208! = 40,320
  • Adjust for repetitions:
    • 40,3203!×2!=40,32012=3,360\frac{40,320}{3! \times 2!} = \frac{40,320}{12} = 3,360

Conclusion: There are 3,360 distinct arrangements of 'NEEDLESS'.

Applying the Formula

  • Identify Repeats: Find repeated elements.
  • Factorials: Use factorial for total and each repeat.
  • Division: Divide total factorial by repeats' factorials.

Example 2: Arranging Letters with Restrictions

Problem: Arrange 'NEEDLESS' so the two S’s are not adjacent.


  • Total Arrangements without Restriction:
    • Formula: 8!3!×2!\frac{8!}{3! \times 2!}
    • Calculation: =40,32012=3,360= \frac{40,320}{12} = 3,360
  • Restricted Arrangements (S's Together):
    • Consider 'SS' as one unit: 7!3! \frac{7!}{3!}
    • Calculation: =5,0406=840= \frac{5,040}{6} = 840
  • Final Calculation (S's Not Together):
    • Total non-restricted - Restricted: 3,360840=2,5203,360 - 840 = 2,520

Conclusion: There are 2,520 ways to arrange 'NEEDLESS' with S’s not adjacent.

Techniques for Repetition

  • Grouping: Treat identical items as one unit.
  • Adjust for Restrictions: Modify the count for constraints.
  • Sequential Approach: Solve in stages.

Example 3: Sequential Password Selection

Problem: Create a password with 3 letters from 'NEEDLESS' followed by 3 numbers (0-9), with number repetition allowed.


  • Letter Selection:
    • Formula: C(n,r)=n!r!(nr)!C(n, r) = \frac{n!}{r!(n-r)!}
    • n=8n = 8, r=3r = 3
    • Letter combinations: C(8,3)=8!3!5!=56C(8, 3) = \frac{8!}{3!5!} = 56
  • Number Selection:
    • Each position can be any of 10 digits (0-9)
    • Number combinations: 103=1,00010^3 = 1,000
  • Combining Results:
    • Total passwords: 56×1,000=56,00056 \times 1,000 = 56,000

Conclusion: There are 56,000 possible passwords.

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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