Number of Arrangements of n Objects of m Types/Examples/2 Types

Example of Use of Number of Arrangements of $n$ Objects of $m$ Types

Let $S$ be a collection of $n$ objects, consisting of:

$p$ objects of one type

$q$ objects of another type.

The total number $N$ of different arrangements of $S$ is given by:

$N = \dfrac {n!} {p! \, q!}$

Proof

An arbitrary arrangement can be made into $p!$ arrangements if the $p$ objects of type $1$ are replaced by $p$ objects that are all different.

Similarly, an arbitrary arrangement can be made into $q!$ arrangements if the $q$ objects of type $2$ are replaced by $q$ objects that are all different.

The total number of arrangements, if all objects are different, would be $n!$

Hence:

$n! = N \times p! \times q!$

Hence the result.

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: Permutations and Combinations: The number of ways of arranging $q$ things in line if $p$ are alike of one kind and $q$ are alike of another kind