# Nakafa Framework: LLM

URL: /en/subject/high-school/12/mathematics/combinatorics/permutation-with-identical-objects
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Output docs content for large language models.

---

export const metadata = {
   title: "Permutation with Identical Objects",
   description: "Learn permutation with identical objects using multinomial formula. Master counting distinct arrangements when items are indistinguishable with examples.",
   authors: [{ name: "Nabil Akbarazzima Fatih" }],
   date: "05/26/2025",
   subject: "Combinatorics",
};

## 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.

## Formula for Identical Object Permutation

For permutation of n objects where there are identical objects, **the formula used** is:

<div className="flex flex-col gap-4">
<BlockMath math="P = \frac{n!}{r_1! \times r_2! \times r_3! \times \cdots \times r_k!}" />
</div>

**Explanation:**

- <InlineMath math="n" /> = total number of objects
- <InlineMath math="r_1, r_2, r_3, \ldots, r_k" /> = number of identical objects in each group
- <InlineMath math="k" /> = 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

Letter identification: K appears 1 time, A appears 3 times (positions 2, 6, 9), L appears 1 time, I appears 1 time, M appears 1 time, N appears 2 times (positions 7, 10), T appears 1 time

**Calculation:**

<div className="flex flex-col gap-4">
<BlockMath math="P = \frac{10!}{1! \times 3! \times 1! \times 1! \times 1! \times 2! \times 1!}" />
<BlockMath math="= \frac{10!}{3! \times 2!}" />
</div>

Simplify the fraction by canceling common factors:

<div className="flex flex-col gap-4">
<BlockMath math="= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3! \times 2!}" />
<BlockMath math="= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{2!}" />
<BlockMath math="= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{2}" />
</div>

Calculate step by step:
- Divide 10 by 2: <InlineMath math="\frac{10}{2} = 5" />
- So: <InlineMath math="5 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4" />
- <InlineMath math="= 5 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 302,400" />

### Example Word PALAPA

For the word "PALAPA" with 6 letters:
Write letters one by one: P-A-L-A-P-A

**Letter identification:** P appears 2 times (positions 1, 5), A appears 3 times (positions 2, 4, 6), L appears 1 time

Calculate each factorial:
- <InlineMath math="6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720" />
- <InlineMath math="2! = 2 \times 1 = 2" />
- <InlineMath math="3! = 3 \times 2 \times 1 = 6" />

So, the calculation is:

<div className="flex flex-col gap-4">
<BlockMath math="P = \frac{6!}{2! \times 3! \times 1!}" />
<BlockMath math="= \frac{6 \times 5 \times 4 \times 3!}{2! \times 3! \times 1!}" />
<BlockMath math="= \frac{6 \times 5 \times 4}{2 \times 1}" />
<BlockMath math="= \frac{6 \times 5 \times 4}{2}" />
</div>

Simplify by dividing 6 by 2:
- <InlineMath math="\frac{6}{2} = 3" />
- So: <InlineMath math="3 \times 5 \times 4 = 60 \text{ arrangements}" />

## 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

### 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: <InlineMath math="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 time
   - A appears 3 times
   - N appears 2 times

4. **Apply formula**

   Use the permutation formula with identical objects:

   <div className="flex flex-col gap-4">
   <BlockMath math="P = \frac{n!}{r_1! \times r_2! \times r_3!}" />
   <BlockMath math="P = \frac{6!}{1! \times 3! \times 2!}" />
   </div>

5. **Calculate factorial**

   Simplify the fraction first:

   <div className="flex flex-col gap-4">
   <BlockMath math="P = \frac{6!}{1! \times 3! \times 2!}" />
   <BlockMath math="= \frac{6 \times 5 \times 4 \times 3!}{1! \times 3! \times 2!}" />
   <BlockMath math="= \frac{6 \times 5 \times 4}{1 \times 2!}" />
   <BlockMath math="= \frac{6 \times 5 \times 4}{2}" />
   </div>

   Calculate with simplification:
   - Divide 6 by 2: <InlineMath math="\frac{6}{2} = 3" />
   - So: <InlineMath math="3 \times 5 \times 4 = 60" />

Therefore, the word "BANANA" can be arranged in **60 different ways**.

## Difference from Regular Permutation

**Regular permutation**: All objects are different, using formula <InlineMath math="n!" />

**Permutation with identical objects**: There are identical objects, using formula:

<div className="flex flex-col gap-4">
<BlockMath math="\frac{n!}{r_1! \times r_2! \times \cdots \times r_k!}" />
</div>

**Comparison example:**

Arranging letters A, B, C, D (all different): <InlineMath math="4! = 24" /> ways

Arranging letters A, A, B, C (some identical): 

<div className="flex flex-col gap-4">
<BlockMath math="\frac{4!}{2!} = \frac{24}{2} = 12 \text{ ways}" />
</div>

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?

### 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 times (positions 1, 5), A appears 3 times (positions 2, 6, 10), T appears 2 times (positions 3, 7), E appears 1 time, I appears 1 time, K appears 1 time
   
   Simplify the fraction by canceling common factors:
   
   <div className="flex flex-col gap-4">
   <BlockMath math="P = \frac{10!}{2! \times 3! \times 2! \times 1! \times 1! \times 1!}" />
   <BlockMath math="= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{2! \times 3! \times 2!}" />
   <BlockMath math="= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{2! \times 2!}" />
   <BlockMath math="= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{4}" />
   </div>

   Calculate with simplification:
   - Complete calculation: <InlineMath math="10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 604,800" />
   - Divide by 4: <InlineMath math="\frac{604,800}{4} = 151,200 \text{ arrangements}" />

2. Total 8 flowers with red 3, white 3, yellow 2
   
   Simplify the fraction by canceling common factors:
   
   <div className="flex flex-col gap-4">
   <BlockMath math="P = \frac{8!}{3! \times 3! \times 2!}" />
   <BlockMath math="= \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3! \times 3! \times 2!}" />
   <BlockMath math="= \frac{8 \times 7 \times 6 \times 5 \times 4}{3! \times 2!}" />
   <BlockMath math="= \frac{8 \times 7 \times 6 \times 5 \times 4}{6 \times 2}" />
   <BlockMath math="= \frac{8 \times 7 \times 6 \times 5 \times 4}{12}" />
   </div>

   Calculate with simplification:
   - Divide 6 by 12: <InlineMath math="\frac{6}{12} = \frac{1}{2}" />
   - So: <InlineMath math="8 \times 7 \times \frac{1}{2} \times 5 \times 4" />
   - <InlineMath math="= 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 times, digit 2 appears 3 times, digit 3 appears 1 time
   
   Simplify the fraction by canceling common factors:
   
   <div className="flex flex-col gap-4">
   <BlockMath math="P = \frac{6!}{2! \times 3! \times 1!}" />
   <BlockMath math="= \frac{6 \times 5 \times 4 \times 3!}{2! \times 3!}" />
   <BlockMath math="= \frac{6 \times 5 \times 4}{2!}" />
   <BlockMath math="= \frac{6 \times 5 \times 4}{2}" />
   </div>

   Calculate with simplification:
   - Divide 6 by 2: <InlineMath math="\frac{6}{2} = 3" />
   - So: <InlineMath math="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 times (positions 1, 8), N appears 2 times (positions 2, 5), D appears 1 time, O appears 1 time, E appears 1 time, S appears 1 time, A appears 1 time
   
   Simplify the fraction by canceling common factors:
   
   <div className="flex flex-col gap-4">
   <BlockMath math="P = \frac{9!}{2! \times 2! \times 1! \times 1! \times 1! \times 1! \times 1!}" />
   <BlockMath math="= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2! \times 2!}" />
   <BlockMath math="= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}{2!}" />
   <BlockMath math="= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}{2}" />
   </div>

   Calculate with simplification:
   - Divide by 2: <InlineMath math="\frac{8}{2} = 4" />
   - So: <InlineMath math="9 \times 4 \times 7 \times 6 \times 5 \times 4 \times 3" />
   - <InlineMath math="= 9 \times 4 \times 7 \times 6 \times 5 \times 4 \times 3 = 90,720 \text{ arrangements}" />