# Permutation

See how to solve a permutation _{n}P_{r}

(+ permutation with repetition, circular permutation, bracelet permutation).

8 examples and their solutions.

## Permutation

### Formula

r numbers

_{n}P_{r}= n⋅(n - 1)⋅(n - 2)⋅ ..._{n}P

_{r}: Starting from n, multiply r numbers.

Meaning: From n things,

pick & arrange r things.

Combination (Math)

_{n}P

_{n}= n! - [1]

_{n}P

_{1}= n - [2]

_{n}P

_{0}= 1 - [3]

[1]

[2]

Starting from n, multiply 1 number.

→ n

→ n

[3]

From n things, pick & arrange 0 things.

→ Don't do anything.

→ 1 way

→ Don't do anything.

→ 1 way

### Example

_{6}P

_{4}

_{6}P

_{4}

= 6⋅5⋅4⋅3 - [1]

= 30⋅12

= 360

[1]

Starting from 6, multiply 4 numbers.

Close

### Example

9 students

Find the number of ways to choose 3 students and arrange them in a row.

Solution Find the number of ways to choose 3 students and arrange them in a row.

_{9}P

_{3}- [1]

= 9⋅8⋅7 - [2]

= 72⋅7

= 504

[1]

From 9 students,

pick & arrange 3 students.

→

pick & arrange 3 students.

→

_{9}P_{3}[2]

Starting from 9, multiply 3 numbers.

Close

### Example

Numbers: 1, 2, 3, 4, 5, 6, 7, 8

By using the numbers once,

find the number of ways to make a 4-digit number.

Solution By using the numbers once,

find the number of ways to make a 4-digit number.

_{8}P

_{4}- [1]

= 8⋅7⋅6⋅5 - [2]

= 56⋅30

= 1680

[1]

From 8 numbers,

pick & arrange 4 numbers.

→

pick & arrange 4 numbers.

→

_{8}P_{4}[2]

Starting from 8, multiply 4 numbers.

Close

### Example

Numbers: 0, 1, 2, 3, 4, 5

By using the numbers once,

find the number of ways to make a 3-digit number.

Solution By using the numbers once,

find the number of ways to make a 3-digit number.

- 0

= 6⋅5⋅4 - 5⋅4 - [2]

= 30⋅4 - 20

= 120 - 20

= 100

_{6}P_{3}-_{5}P_{2}- [1]= 6⋅5⋅4 - 5⋅4 - [2]

= 30⋅4 - 20

= 120 - 20

= 100

[1]

6 numbers

Make a 3-digit number.

→ From 6 numbers,

pick & arrange 3 numbers.

→

_{6}P

_{3}

- 0

But, if the hundredth number is 0,

it's not a 3-digit number.

(012, 023, ...)

So subtract these number of ways.

→ From 5 numbers (1, 2, 3, 4, 5),

pick & arrange 2 numbers.

→

_{5}P

_{2}

[2]

_{6}P

_{3}

Starting from 6, multiply 3 numbers.

_{5}P

_{2}

Starting from 5, multiply 2 numbers.

Close

## Permutation with Repetition

### Formula

N = n!p!⋅q!⋅r!⋅...

n = p + q + r + ... p, q, r: Numbers of identical things

### Example

Letters: a, a, a, b, b, c, c

By using each letter once,

find the number of ways to make a 7-letter word.

Solution By using each letter once,

find the number of ways to make a 7-letter word.

n = 7

a: 3

b: 2

c: 2

N = 7!3!⋅2!⋅2!

= 7⋅6⋅5⋅4⋅3!3!⋅2⋅1⋅2⋅1 - [1]

= 7⋅6⋅5

= 7⋅30

= 210

a: 3

b: 2

c: 2

N = 7!3!⋅2!⋅2!

= 7⋅6⋅5⋅4⋅3!3!⋅2⋅1⋅2⋅1 - [1]

= 7⋅6⋅5

= 7⋅30

= 210

[1]

Close

### Example

Find the number of shortest paths to move from A to B.

Solution →: 3

↓: 3 - [1]

n = 6

N = 6!3!⋅3!

= 6⋅5⋅4⋅3!3!⋅3⋅2⋅1

= 5⋅4

= 20

↓: 3 - [1]

n = 6

N = 6!3!⋅3!

= 6⋅5⋅4⋅3!3!⋅3⋅2⋅1

= 5⋅4

= 20

[1]

To make the shortest path,

you should move either → or ↓.

Width: 3 blocks

Height: 3 blocks

→ Pick & arrange →, →, →, ↓, ↓, ↓.

you should move either → or ↓.

Width: 3 blocks

Height: 3 blocks

→ Pick & arrange →, →, →, ↓, ↓, ↓.

Close

## Circular Permutation

### Definition

when arranging things in a circle.

These two cases are the same case

because when you rotate the left case ↷,

you can get the right case.

### Formula

N = (n - 1)!

### Example

5 students

Find the number of ways to make them sit at a round table.

Solution Find the number of ways to make them sit at a round table.

n = 5

N = 4! - [1]

= 4⋅3⋅2⋅1 - [2]

= 12⋅2

= 24

N = 4! - [1]

= 4⋅3⋅2⋅1 - [2]

= 12⋅2

= 24

[1]

Sitting in a round table

→ Circular permutation

→ Circular permutation

[2]

Close

## Bracelet Permutation

### Definition

when arranging things in a bracelet (or a necklace).

These two cases are the same case

because when you flip the left case,

you can get the right case.

### Formula

N = (n - 1)!2

### Example

7 beads

Find the number of ways to make a bracelet by using these beads.

Solution Find the number of ways to make a bracelet by using these beads.

n = 7

N = 6!2 - [1]

= 6⋅5⋅4⋅3⋅2⋅12 - [2]

= 6⋅5⋅4⋅3

= 30⋅12

= 360

N = 6!2 - [1]

= 6⋅5⋅4⋅3⋅2⋅12 - [2]

= 6⋅5⋅4⋅3

= 30⋅12

= 360

[1]

Making a bracelet

→ Bracelet permutation

→ Bracelet permutation

[2]

Close