Permutations with Repetition

Permutations with Repetition

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

Formula

n!/(p!q!r!...) n: Total number of things. p, q, r, ... : Numbers of identical things

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

Letters {a, a, b, c, d} are given. By using each letter once, find the number of ways to make a 5-letter word.

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

Letters {a, a, a, b, b, c, c} are given. By using each letter once, find the number of ways to make a 7-letter word.

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.