# Nakafa Framework: LLM

URL: /en/subject/high-school/12/mathematics/combinatorics/permutation-of-n-items-from-n-objects
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---

export const metadata = {
   title: "Permutation of n Items from n Objects",
   description: "Master permutation of n items from n objects using factorial formulas. Learn complete arrangements with step-by-step examples and practical exercises.",
   authors: [{ name: "Nabil Akbarazzima Fatih" }],
   date: "05/26/2025",
   subject: "Combinatorics",
};

## Understanding Permutation of n Items from n Objects

**Permutation of n items from n objects** is an arrangement of all available objects where each object is used exactly once and **the order of arrangement is very important**. In this case, we use all available objects without any remaining.

Imagine arranging a photo line for all students in a class. If there are 5 students, then we will arrange all five students in one line. Each student will get one position, none will sit on the backup bench, and Andi's position in front or back gives different results.

## Complete Permutation Formula

For permutation of n items from n objects, **the formula used** is:

<div className="flex flex-col gap-4">
<BlockMath math="nP_n = \frac{n!}{(n-n)!} = \frac{n!}{0!}" />
</div>

**Explanation of why this formula becomes simple:**
- <InlineMath math="(n-n) = 0" />, so the denominator becomes <InlineMath math="0!" />
- Based on mathematical definition, <InlineMath math="0! = 1" />
- Therefore: <InlineMath math="\frac{n!}{0!} = \frac{n!}{1} = n!" />

So, the formula is:

<div className="flex flex-col gap-4">
<BlockMath math="nP_n = n!" />
</div>

## Factorial Concept

**Factorial** is the consecutive multiplication of positive integers. Factorial of number n is written as <InlineMath math="n!" /> and defined as:

<div className="flex flex-col gap-4">
<BlockMath math="n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1" />
</div>

**Special rules** for factorial:
- <InlineMath math="0! = 1" /> (based on mathematical definition)
- <InlineMath math="1! = 1" />

## Application in Daily Life

### School Organization

Suppose there are 4 students who will fill 4 positions in class leadership: president, vice president, secretary, and treasurer. Each student can only hold one position.

The number of ways to arrange the leadership is:

<div className="flex flex-col gap-4">
<BlockMath math="4P_4 = 4! = 4 \times 3 \times 2 \times 1 = 24 \text{ ways}" />
</div>

### Seating Arrangement

A family consisting of 6 members will sit in a row on a sofa for a family photo. The number of ways they can be arranged is:

<div className="flex flex-col gap-4">
<BlockMath math="6P_6 = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \text{ ways}" />
</div>

## Systematic Calculation Steps

To calculate permutation of n items from n objects, follow these steps:

1. **Identify the number of objects**: Ensure all objects will be used
2. **Apply the formula**: Use <InlineMath math="n!" />
3. **Calculate factorial**: Multiply consecutively from n to 1
4. **Verify the result**: Ensure the calculation is correct

### Detailed Calculation

Calculating <InlineMath math="5!" /> with clear steps:

<div className="flex flex-col gap-4">
<BlockMath math="5! = 5 \times 4 \times 3 \times 2 \times 1" />
<BlockMath math="= 20 \times 3 \times 2 \times 1" />
<BlockMath math="= 60 \times 2 \times 1" />
<BlockMath math="= 120 \times 1 = 120" />
</div>

## Difference from Partial Permutation

**Complete permutation (n from n objects)**: Uses all available objects.
**Partial permutation (r from n objects)**: Only uses some objects.

**Concrete example**: 
- Complete permutation: Arranging 5 books in 5 positions on a shelf = <InlineMath math="5!" /> = 120 ways
- Partial permutation: Selecting and arranging 3 books from 5 available books = <InlineMath math="5P_3" /> = 60 ways

**Detailed calculation for partial permutation**:

<div className="flex flex-col gap-4">
<BlockMath math="5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!}" />
<BlockMath math="= \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}" />
<BlockMath math="= \frac{120}{2} = 60 \text{ ways}" />
</div>

In permutation of n items from n objects, **no objects are left over** and **all positions must be filled**.

## Exercises

1. A photography team wants to arrange 7 models for a photo session in one line. How many different ways can they arrange the seven models?

2. In a running competition, there are 5 participants who must all finish. How many different finishing order possibilities are there?

3. A chef wants to arrange 6 different types of food on a table in one straight line. How many different ways can he arrange the food?

4. A library has 8 different books that will be arranged on one shelf. If all books must be placed on that shelf, how many possible arrangements are there?

### Answer Key

1. Given: 7 models will be arranged in one line (all models used)
   <div className="flex flex-col gap-4">
   <BlockMath math="7P_7 = 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \text{ ways}" />
   </div>

2. Given: 5 running participants with different finishing order (all participants finish)
   <div className="flex flex-col gap-4">
   <BlockMath math="5P_5 = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \text{ possible orders}" />
   </div>

3. Given: 6 types of food will be arranged in one straight line (all food arranged)
   <div className="flex flex-col gap-4">
   <BlockMath math="6P_6 = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \text{ ways}" />
   </div>

4. Given: 8 different books will be arranged on one shelf (all books arranged)
   <div className="flex flex-col gap-4">
   <BlockMath math="8P_8 = 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \text{ ways}" />
   </div>