# Permutation: Repetition

How to solve a permutation with repetition: formula, 3 examples, and their solutions.

## Formula

If there's p of idential things,

q of another idential things,

r of another idential things,

...

then the number of ways

to arrange all of them is

n!/[p!⋅q!⋅r!⋅...].

n: Total number of things

This is the permutation with repetition formula.

## ExampleWord Problem

There are 5 letters.

n = 5

There are 2 of a-s.

To make a 5-letter word,

arrange these 5 letters in a row.

So the number of ways

to make a 5-letter word is,

5!/2!.

5! = 5⋅4⋅3⋅2!

Factorial

Cancel both 2!.

5⋅4 = 20

20⋅3 = 60

So 60 is the answer.

## ExampleWord Problem

There are 7 letters.

n = 7

There are

3 of a-s,

2 of b-s,

and 2 of c-s.

To make a 7-letter word,

arrange these 7 letters in a row.

So the number of ways

to make a 7-letter word is,

7!/[3!⋅2!⋅2!].

7! = 7⋅6⋅5⋅4⋅3!

2! = 2⋅1 = 2

2! = 2⋅1 = 2

Cancel both 3!.

Cancel 4 in the numerator

and cancel 2⋅2 in the denominator.

6⋅5 = 30

7⋅30 = 210

So 210 is the answer.

## ExampleNumber of Shortest Paths

In the given map,

there's no disconnected part.

So, instead of adding the paths,

you can solve this by using

permutation with repetition formula.

To make the shortest path,

you should move

→ (left) 3 times

and

↓ (downward) 3 times.

To make a shortest path,

arrange these 6 directions in a row.

So the number of the shortest paths is

6!/[3!⋅3!].

6! = 6⋅5⋅4⋅3!

3! = 3⋅2⋅1 = 3⋅2

Cancel both 3!.

Cancel 6 in the numerator

and cancel 3⋅2 in the denominator.

5⋅4 = 20

So 20 is the answer.