Permutations with Groups of Equal Elements

Topics: Permutation

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Let’s say we have a set of $n$ elements with group of equal elements. We have $n_1$ elements of type 1, $n_2$ elements of type 2, etc. ($n=n_1+n_2+\cdots +n_k$).

In such a case, the number of distinguishable permutations we can form with the $n$ elements is:

$${}_n{\text{P}}_{n_1,\dots ,n_k}=\frac{n!}{n_1!n_2!\dots n_k!}$$

For a set whose elements are all distinct, see permutations with no repetitions.

Example

How many permutations can be made with the letters of the word “PEPPER”?

Notice that “PEPPER” has 6 letters:

• $3$ Ps
• $2$ Es
• $1$ R

Thus, we can have ${}_6{\text{P}}_3,2,1=\frac{6!}{3!2!1!}=\frac{6(5)4}{2}=60$ permutations.