Permutation of All Objects

Understanding Permutation of All Objects

Permutation of all objects is an arrangement of all available objects, namely $n$ items from $n$ 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 $5 \text{ students}$. All five students are placed in one line. Each student gets one position, none sits 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:

Explanation of why this formula becomes simple:

• $(n-n) = 0$, so the denominator becomes $0!$
• Based on mathematical definition, $0! = 1$
• Therefore: $\frac{n!}{0!} = \frac{n!}{1} = n!$

So, the formula is:

Factorial Concept

Factorial is the consecutive multiplication of positive integers. The factorial of a number $n$ is written as $n!$ and defined as:

Special rules for factorial:

• $0! = 1$ (based on mathematical definition)
• $1! = 1$

Application in Daily Life

School Organization

Suppose there are $4 \text{ students}$ who will fill $4$ positions in a student committee: president, vice president, secretary, and treasurer. Each student can only hold one position.

The number of ways to arrange the leadership is:

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:

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 $n!$
3. Calculate factorial: Multiply consecutively from $n$ to $1$
4. Verify the result: Ensure the calculation is correct

Detailed Calculation

Calculating $5!$ with clear steps:

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 $5 \text{ books}$ in $5 \text{ positions}$ on a shelf is $5! = 120 \text{ ways}$
• Partial permutation: Selecting and arranging $3 \text{ books}$ from $5 \text{ available books}$ is $5P_3 = 60 \text{ ways}$

Detailed calculation for partial permutation:

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 $7 \text{ models}$ for a photo session in one line. How many different ways can they arrange the seven models?

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

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

4. A library has $8 \text{ different books}$ 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: $7 \text{ models}$ will be arranged in one line (all models used)

   $$7P_7 = 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \text{ ways}$$

2. Given: $5 \text{ running participants}$ with different finishing order (all participants finish)

   $$5P_5 = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \text{ possible orders}$$

3. Given: $6 \text{ types of food}$ will be arranged in one straight line (all food arranged)

   $$6P_6 = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \text{ ways}$$

4. Given: $8 \text{ different books}$ will be arranged on one shelf (all books arranged)

   $$8P_8 = 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \text{ ways}$$