# Nakafa Learning Content
> For AI agents: use [llms.txt](https://nakafa.com/llms.txt) for the site index. Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`.
URL: https://nakafa.com/en/subjects/mathematics/combinatorics/permutation-with-identical-objects
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/mathematics/combinatorics/permutation-with-identical-objects/en.mdx
Learn permutation with identical objects using multinomial formula. Learn counting distinct arrangements when items are indistinguishable with examples.
---
## Understanding Permutation with Identical Objects
**Permutation with identical objects** is an arrangement of objects where there are several objects that are identical or the same. When there are identical objects, the number of different arrangements will decrease because **exchanging identical objects does not produce new arrangements**.
Imagine arranging letters from the word "MAMA". Although there are $$4$$ letters, we cannot distinguish between the first $$M$$ and the second $$M$$, or the first $$A$$ and the second $$A$$. As a result, arrangements that look different but use the same letters in different positions are considered identical.
Visible text: Imagine arranging letters from the word "MAMA". Although there are letters, we cannot distinguish between the first and the second , or the first and the second . As a result, arrangements that look different but use the same letters in different positions are considered identical.
## Formula for Identical Object Permutation
For permutation of $$n$$ objects where there are identical objects, **the formula used** is:
Visible text: For permutation of objects where there are identical objects, **the formula used** is:
Component: MathContainer
Children:
```math
P = \frac{n!}{r_1! \times r_2! \times r_3! \times \cdots \times r_k!}
```
**Explanation:**
- $$n$$ = total number of objects
- $$r_1, r_2, r_3, \ldots, r_k$$ = number of identical objects in each group
- $$k$$ = number of groups of identical objects
Visible text: - = total number of objects
- = number of identical objects in each group
- = number of groups of identical objects
**How to identify identical objects**: Count how many times each object appears in the entire arrangement, not just looking at different objects.
## Application to Words and Letters
### Example Word KALIMANTAN
Let's calculate how many letter arrangements can be made from the word "KALIMANTAN".
**Systematic identification steps:**
Write letters one by one: K-A-L-I-M-A-N-T-A-N
Total letters: $$10$$
Visible text: Total letters:
Letter identification: $$K$$ appears $$1 \text{ time}$$, $$A$$ appears $$3 \text{ times}$$ (positions $$2, 6, 9$$), $$L$$ appears $$1 \text{ time}$$, $$I$$ appears $$1 \text{ time}$$, $$M$$ appears $$1 \text{ time}$$, $$N$$ appears $$2 \text{ times}$$ (positions $$7, 10$$), and $$T$$ appears $$1 \text{ time}$$
Visible text: Letter identification: appears , appears (positions ), appears , appears , appears , appears (positions ), and appears
**Calculation:**
Component: MathContainer
Children:
```math
P = \frac{10!}{1! \times 3! \times 1! \times 1! \times 1! \times 2! \times 1!}
```
```math
= \frac{10!}{3! \times 2!}
```
Simplify the fraction by canceling common factors:
Component: MathContainer
Children:
```math
= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3! \times 2!}
```
```math
= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{2!}
```
```math
= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{2}
```
Calculate step by step:
- Divide $$10$$ by $$2$$: $$\frac{10}{2} = 5$$
- So: $$5 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4$$
- $$= 5 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 302{,}400$$
Visible text: - Divide by :
- So:
-
### Example Word PALAPA
For the word "PALAPA" with $$6$$ letters:
Write letters one by one: P-A-L-A-P-A
Visible text: For the word "PALAPA" with letters:
Write letters one by one: P-A-L-A-P-A
**Letter identification:** $$P$$ appears $$2 \text{ times}$$ (positions $$1, 5$$), $$A$$ appears $$3 \text{ times}$$ (positions $$2, 4, 6$$), and $$L$$ appears $$1 \text{ time}$$
Visible text: **Letter identification:** appears (positions ), appears (positions ), and appears
Calculate each factorial:
- $$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$$
- $$2! = 2 \times 1 = 2$$
- $$3! = 3 \times 2 \times 1 = 6$$
Visible text: -
-
-
So, the calculation is:
Component: MathContainer
Children:
```math
P = \frac{6!}{2! \times 3! \times 1!}
```
```math
= \frac{6 \times 5 \times 4 \times 3!}{2! \times 3! \times 1!}
```
```math
= \frac{6 \times 5 \times 4}{2 \times 1}
```
```math
= \frac{6 \times 5 \times 4}{2}
```
Simplify by dividing $$6$$ by $$2$$:
Visible text: Simplify by dividing by :
- $$\frac{6}{2} = 3$$
- So: $$3 \times 5 \times 4 = 60 \text{ arrangements}$$
Visible text: -
- So:
## Systematic Calculation Steps
To solve permutation problems with identical objects, follow these steps:
1. **Count total objects**: Determine the value of $$n$$
2. **Identify identical objects**: Group objects that are identical
3. **Count frequency**: Determine how many times each object appears
4. **Apply formula**: Insert into the permutation formula
5. **Calculate factorial**: Complete the calculation carefully
Visible text: 1. **Count total objects**: Determine the value of
2. **Identify identical objects**: Group objects that are identical
3. **Count frequency**: Determine how many times each object appears
4. **Apply formula**: Insert into the permutation formula
5. **Calculate factorial**: Complete the calculation carefully
### Word BANANA
Let's apply these steps to find arrangements of the word "BANANA":
1. **Count total objects**
Write letters one by one: B-A-N-A-N-A
Total letters: $$n = 6$$
2. **Identify identical objects**
Group identical letters together:
- B group: B
- A group: A, A, A
- N group: N, N
3. **Count frequency**
Count how many times each letter appears:
- $$B$$ appears $$1 \text{ time}$$
- A appears $$3 \text{ times}$$
- N appears $$2 \text{ times}$$
4. **Apply formula**
Use the permutation formula with identical objects:
```math
P = \frac{n!}{r_1! \times r_2! \times r_3!}
```
```math
P = \frac{6!}{1! \times 3! \times 2!}
```
5. **Calculate factorial**
Simplify the fraction first:
```math
P = \frac{6!}{1! \times 3! \times 2!}
```
```math
= \frac{6 \times 5 \times 4 \times 3!}{1! \times 3! \times 2!}
```
```math
= \frac{6 \times 5 \times 4}{1 \times 2!}
```
```math
= \frac{6 \times 5 \times 4}{2}
```
Calculate with simplification:
- Divide $$6$$ by $$2$$: $$\frac{6}{2} = 3$$
- So: $$3 \times 5 \times 4 = 60$$
Visible text: 1. **Count total objects**
Write letters one by one: B-A-N-A-N-A
Total letters:
2. **Identify identical objects**
Group identical letters together:
- B group: B
- A group: A, A, A
- N group: N, N
3. **Count frequency**
Count how many times each letter appears:
- appears
- A appears
- N appears
4. **Apply formula**
Use the permutation formula with identical objects:
5. **Calculate factorial**
Simplify the fraction first:
Calculate with simplification:
- Divide by :
- So:
Therefore, the word "BANANA" can be arranged in **$$60$$ different ways**.
Visible text: Therefore, the word "BANANA" can be arranged in ** different ways**.
## Difference from Regular Permutation
**Regular permutation**: All objects are different, using formula $$n!$$
Visible text: **Regular permutation**: All objects are different, using formula
**Permutation with identical objects**: There are identical objects, using formula:
Component: MathContainer
Children:
```math
\frac{n!}{r_1! \times r_2! \times \cdots \times r_k!}
```
**Comparison example:**
Arranging letters A, B, C, D (all different): $$4! = 24 \text{ ways}$$
Visible text: Arranging letters A, B, C, D (all different):
Arranging letters $$A, A, B, C$$ (some identical):
Visible text: Arranging letters (some identical):
Component: MathContainer
Children:
```math
\frac{4!}{2!} = \frac{24}{2} = 12 \text{ ways}
```
Identical objects **reduce** the number of arrangements because exchanging identical objects does not produce differences.
## Exercises
1. How many letter arrangements can be made from the word "MATEMATIKA"?
2. A flower shop has $$8$$ roses where $$3$$ are red, $$3$$ are white, and $$2$$ are yellow. How many ways can these flowers be arranged in a row?
3. From the digits $$1, 1, 2, 2, 2, 3$$, how many $$6$$-digit numbers can be formed?
4. How many different letter arrangements does the word "INDONESIA" have?
Visible text: 1. How many letter arrangements can be made from the word "MATEMATIKA"?
2. A flower shop has roses where are red, are white, and are yellow. How many ways can these flowers be arranged in a row?
3. From the digits , how many -digit numbers can be formed?
4. How many different letter arrangements does the word "INDONESIA" have?
### Answer Key
1. The word "MATEMATIKA" has $$10$$ letters
**Letters one by one:** M-A-T-E-M-A-T-I-K-A
**Letter identification:** $$M$$ appears $$2 \text{ times}$$ (positions $$1, 5$$), $$A$$ appears $$3 \text{ times}$$ (positions $$2, 6, 10$$), $$T$$ appears $$2 \text{ times}$$ (positions $$3, 7$$), $$E$$ appears $$1 \text{ time}$$, $$I$$ appears $$1 \text{ time}$$, and $$K$$ appears $$1 \text{ time}$$
Simplify the fraction by canceling common factors:
```math
P = \frac{10!}{2! \times 3! \times 2! \times 1! \times 1! \times 1!}
```
```math
= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{2! \times 3! \times 2!}
```
```math
= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{2! \times 2!}
```
```math
= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{4}
```
Calculate with simplification:
- Complete calculation: $$10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 604{,}800$$
- Divide by $$4$$: $$\frac{604{,}800}{4} = 151{,}200 \text{ arrangements}$$
2. Total $$8$$ flowers with red $$3$$, white $$3$$, and yellow $$2$$
Simplify the fraction by canceling common factors:
```math
P = \frac{8!}{3! \times 3! \times 2!}
```
```math
= \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3! \times 3! \times 2!}
```
```math
= \frac{8 \times 7 \times 6 \times 5 \times 4}{3! \times 2!}
```
```math
= \frac{8 \times 7 \times 6 \times 5 \times 4}{6 \times 2}
```
```math
= \frac{8 \times 7 \times 6 \times 5 \times 4}{12}
```
Calculate with simplification:
- Divide $$6$$ by $$12$$: $$\frac{6}{12} = \frac{1}{2}$$
- So: $$8 \times 7 \times \frac{1}{2} \times 5 \times 4$$
- $$= 4 \times 7 \times 5 \times 4 = 560 \text{ ways}$$
3. Digits $$1, 1, 2, 2, 2, 3$$ (total $$6$$ digits)
**Digit identification:** digit $$1$$ appears $$2 \text{ times}$$, digit $$2$$ appears $$3 \text{ times}$$, and digit $$3$$ appears $$1 \text{ time}$$
Simplify the fraction by canceling common factors:
```math
P = \frac{6!}{2! \times 3! \times 1!}
```
```math
= \frac{6 \times 5 \times 4 \times 3!}{2! \times 3!}
```
```math
= \frac{6 \times 5 \times 4}{2!}
```
```math
= \frac{6 \times 5 \times 4}{2}
```
Calculate with simplification:
- Divide $$6$$ by $$2$$: $$\frac{6}{2} = 3$$
- So: $$3 \times 5 \times 4 = 60 \text{ numbers}$$
4. The word "INDONESIA" has $$9$$ letters
**Letters one by one:** I-N-D-O-N-E-S-I-A
**Letter identification:** $$I$$ appears $$2 \text{ times}$$ (positions $$1, 8$$), $$N$$ appears $$2 \text{ times}$$ (positions $$2, 5$$), $$D$$ appears $$1 \text{ time}$$, $$O$$ appears $$1 \text{ time}$$, $$E$$ appears $$1 \text{ time}$$, $$S$$ appears $$1 \text{ time}$$, and $$A$$ appears $$1 \text{ time}$$
Simplify the fraction by canceling common factors:
```math
P = \frac{9!}{2! \times 2! \times 1! \times 1! \times 1! \times 1! \times 1!}
```
```math
= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2! \times 2!}
```
```math
= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}{2!}
```
```math
= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}{2}
```
Calculate with simplification:
- Divide by $$2$$: $$\frac{8}{2} = 4$$
- So: $$9 \times 4 \times 7 \times 6 \times 5 \times 4 \times 3$$
- $$= 9 \times 4 \times 7 \times 6 \times 5 \times 4 \times 3 = 90{,}720 \text{ arrangements}$$
Visible text: 1. The word "MATEMATIKA" has letters
**Letters one by one:** M-A-T-E-M-A-T-I-K-A
**Letter identification:** appears (positions ), appears (positions ), appears (positions ), appears , appears , and appears
Simplify the fraction by canceling common factors:
Calculate with simplification:
- Complete calculation:
- Divide by :
2. Total flowers with red , white , and yellow
Simplify the fraction by canceling common factors:
Calculate with simplification:
- Divide by :
- So:
-
3. Digits (total digits)
**Digit identification:** digit appears , digit appears , and digit appears
Simplify the fraction by canceling common factors:
Calculate with simplification:
- Divide by :
- So:
4. The word "INDONESIA" has letters
**Letters one by one:** I-N-D-O-N-E-S-I-A
**Letter identification:** appears (positions ), appears (positions ), appears , appears , appears , appears , and appears
Simplify the fraction by canceling common factors:
Calculate with simplification:
- Divide by :
- So:
-