# Factorial

See how to solve a factorial n!

(+word examples).

4 examples and their solutions.

## Factorial

### Formula

n! = n⋅(n - 1)⋅(n - 2)⋅ ... ⋅3⋅2⋅1

Multiply from n to 1. Meaning: Pick & arrange n things

1! = 1

0! = 1

0! is defined as 1. 0! = 1

### Example

5!

Solution 5! = 5⋅4⋅3⋅2⋅1

= 20⋅6

= 120

= 20⋅6

= 120

Close

### Example

7!4!

Solution 7!4!

= 7⋅6⋅5⋅4!4!

= 7⋅6⋅5 - [1]

= 7⋅30

= 210

= 7⋅6⋅5⋅4!4!

= 7⋅6⋅5 - [1]

= 7⋅30

= 210

[1]

7! = 7⋅6⋅5⋅4⋅3⋅2⋅1

= 7⋅6⋅5⋅4!

To cancel the denominator 4! easily,

7! = 7⋅6⋅5⋅4!.

= 7⋅6⋅5⋅4!

To cancel the denominator 4! easily,

7! = 7⋅6⋅5⋅4!.

Close

### Example

4 letters: a, b, c, d

Find the number of ways to arrange the letters.

Solution Find the number of ways to arrange the letters.

a, b, c, d

N = 4! - [1]

= 4⋅3⋅2⋅1

= 4⋅6

= 24

N = 4! - [1]

= 4⋅3⋅2⋅1

= 4⋅6

= 24

[1]

Pick and arrange 4 letters.

→ 4!

→ 4!

Close

### Example

3 letters: a, b, c

4 numbers: 1, 2, 3, 4

Find the number of ways to arrange the letters and the numbers

when all letters are adjacent to each other.

Solution 4 numbers: 1, 2, 3, 4

Find the number of ways to arrange the letters and the numbers

when all letters are adjacent to each other.

a, b, c, 1, 2, 3, 4 - [1]

N = 5!⋅3! - [2]

= 5⋅4⋅3⋅2⋅1⋅32⋅1

= 20⋅6⋅6

= 20⋅36

= 720

N = 5!⋅3! - [2]

= 5⋅4⋅3⋅2⋅1⋅32⋅1

= 20⋅6⋅6

= 20⋅36

= 720

[1]

All letters are adjacent to each other.

→ Group the letters.

→ Group the letters.

[2]

Group, 1, 2, 3, 4

→ Pick and arrange 5 things.

→ 5!

For each case,

the letters (a, b, c) can be arranged in the Group.

→ Pick and arrange 3 things.

→ × 3!

Number of Ways (Math)

→ Pick and arrange 5 things.

→ 5!

For each case,

the letters (a, b, c) can be arranged in the Group.

→ Pick and arrange 3 things.

→ × 3!

Number of Ways (Math)

Close