# Number of Arrangements of n Objects of m Types

## Theorem

Let $S$ be a collection of $n$ objects.

Let these $n$ objects be of $m$ different types, as follows:

Let there be:

such that:

- for each $j \in \set {1, 2, \ldots, m}$, all objects of type $j$ are indistinguishable from each other
- $k_1 + k_2 + \cdots + k_m = n$

Then the total number $N$ of different arrangements of $S$ is given by the multinomial coefficient:

- $N = \dbinom n {k_1, k_2, \ldots, k_m} = \dfrac {n!} {k_1! \, k_2! \cdots k_m!}$

## Proof

Let $N$ be the number of different arrangements of $S$.

First suppose that all $n$ objects are distinct one from another.

Then from Number of Permutations of All Elements:

- $N = n!$

Now suppose that $k_j$ elements of $S$ are indistinguishable from each other.

From Number of Permutations of All Elements, there are $k_j!$ different arrangements of those $k_j$ elements.

Hence the $n!$ arrangements of $S$ can each be grouped into $\dfrac {n!} {k_j!}$ partitions, each with $k_j$ elements, such that all the arrangements in each partition are indistinguishable from each other.

This applies for all $m$ types of objects.

The result follows.

$\blacksquare$

Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: This is intuitive and not at all rigorous. I throw it open for anyone either to improve this informal demonstration or write a further proof along rigorous lines.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Examples

### $2$ Types

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

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

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

### $3$ Types

Let $S$ be a collection of $\paren {p + q + r}$ objects.

Let $S$ need to be partitioned into $3$ subsets of size $p$, $q$ and $r$ such that $p \ne q$, $q \ne r$ and $r \ne p$.

The total number $N$ of ways this can be done is:

- $N = \dfrac {\paren {p + q + r}!} {p! \, q! \, r!}$

### Set of $3 p$ Objects of $3$ Equal Sized Subsets

Let $S$ be a collection of $3 p$ objects.

Let $S$ need to be partitioned into $3$ subsets of size $p$.

The total number $N$ of ways this can be done is:

- $N = \dfrac {\paren {3 p}!} {\paren {p!}^3 \times 3!}$

### Letters in *added*

Let $N$ be the number of different arrangements of the letters in the word $\texttt{added}$.

Then:

- $N = 20$

### $6$ people in $3$ pairs

Let $N$ be the number of ways $6$ people can be partitioned into $3$ (unordered) pairs.

Then:

- $N = 15$

### $10$ people in $3$ groups of sizes $5$, $3$ and $2$

Let $N$ be the number of ways $10$ people can be partitioned into $3$ sets: one with $5$, one with $3$ and one with $2$ people.

Then:

- $N = 2520$