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: $\text{Arrangements} = \frac{n!}{n_1! \times n_2! \times \ldots \times n_k!}$
  • $n$: Total objects.
  • $n_1, n_2, \ldots, n_k$: Repeats of each object.

Example 1: Word Arrangements

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

Solution:

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

Calculation:

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

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.

Solution:

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

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.

Solution:

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

Conclusion: There are 56,000 possible passwords.