Permutation of All Objects

Assigning Every Object a Place in the Order

Permutation of all objects arranges all $n$ available objects, uses each object exactly once, and treats different orders as different arrangements.

Five students are lining up for a photo. All five receive one position, and every different order counts as a different arrangement.

Complete Permutation Formula

Set $r=n$ in the general permutation formula:

Since all objects are used:

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

Because $0!=1$, a complete permutation has the formula:

Factorial Definition

The factorial $n!$ is the product of the positive integers from $n$ down to $1$:

By definition:

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

Assigning All Objects to Distinct Positions

Each arrangement below gives every object its own position, so no object is left out and none repeats. Counting starts with the number of objects and multiplies one choice fewer at every step.

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:

Calculating Complete Permutations

For a complete permutation of $n$ objects:

1. Count the objects: Confirm that every object is used
2. Write the formula: Use $n!$
3. Expand the factorial: Multiply from $n$ down to $1$
4. Check the arithmetic: Recalculate the product

Expanding Five Factorial

Expand $5!$ one factor at a time:

Complete and Partial Permutations

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

For five books:

• 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

All three problems arrange every object in a line and ask for the number of orders, so each one uses the same factorial count.

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 $6$ 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)

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