 # 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  letters.

 of a-s are identical.

So the number of ways
to arrange these 5 letters are
! / !.

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

## Example 2 There are  letters.

 of a-s are identical.
 of b-s are identical.
And  of c-s are identical.

So the number of ways
to arrange these 7 letters are
! / ( !⋅!⋅! ).

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