# Permutations with Repetition

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

## Formula

Assume that

for a sample space that has *n* things,

there are *p* of identical things,*q* of identical things,*r* of identical things,

... .

Then the number of ways

to arrange these things are*n*! / (*p*!*q*!*r*! ...).

## Example 1

There are [5] letters.

[2] of a-s are identical.

So the number of ways

to arrange these 5 letters are

[5]! / [2]!.

5! is

multiplying from 5 to 1.

But, instead of writing 5⋅4⋅3⋅[2⋅1] / 2!,

write 5⋅4⋅3⋅[2!] / 2!,

in order to solve this easier.

Factorial (*n*!) - Example 2

Cancel 2!

in both of the numerator and the denominator.

Then (given) = 5⋅4⋅3.

5⋅4 = 20

20⋅3 = 60

So 60 is the answer.

## Example 2

There are [7] letters.

[3] of a-s are identical.

[2] of b-s are identical.

And [2] of c-s are identical.

So the number of ways

to arrange these 7 letters are

[7]! / ( [3]!⋅[2]!⋅[2]! ).

7! = 7⋅6⋅5⋅4⋅[3!]

Leave 3!.

And change 2! to 2⋅1.

Cancel 3!

in both of the numerator and the denominator.

Cancel 4 in the numerator

and cancel 2⋅1⋅2⋅1.

Then (given) = 7⋅6⋅5.

6⋅5 = 30

7⋅30 = 210

So 210 is the answer.