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.

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.