Understanding the Multinomial Coefficient

June 16, 2026

The multinomial coefficient is defined as the number of ways can we partition the set $\{1, 2, \ldots, n\}$ into groups of size $k_1, k_2, \ldots, k_r$. It is written as follows:

\[\binom{n}{k_1,\, k_2,\, \ldots,\, k_r}\]

For example, $\binom{4}{2, 2}$ is the number of ways to partition the numbers $1, 2, 3, 4$ into groups of size $2$ and $2$. It turns out there are $3$ ways to do that¹:

$\{1, 2\} | \{3, 4\}$
$\{1, 3\} | \{2, 4\}$
$\{1, 4\} | \{2, 3\}$

How do we get $3$? In this small example, it was easy enough for us to write out all the partitions. But what if $n$ is large? How can we derive the multinomial coefficient for ourselves?

Deriving it for ourselves

Well, consider this—we’re going to start with something easier. Let’s consider all the permutations of $\{1, 2, 3, 4\}$ e.g.

$(1, 2, 3, 4)$
$(1, 2, 4, 3)$
$(1, 3, 2, 4)$
$(1, 3, 4, 2)$
$(1, 4, 2, 3)$
$(1, 4, 3, 2)$
$(2, 1, 3, 4)$
$\vdots$
$(4, 3, 2, 1)$

There are $n! = 4! = 24$ of these permuations.

Now, let’s think about grouping these permutations together into clusters, where every cluster is going to correspond to all the permutations of a particular partitioning.

For example, consider the partition $\{1, 2\} | \{3, 4\}$. All of its “permutations” are:

$(1, 2, 3, 4)$
$(1, 2, 4, 3)$
$(2, 1, 3, 4)$
$(2, 1, 4, 3)$

So we see there are $4$ ways to permute this partitioning.

Now imagine repeating this process for each possible partitioning. We’d have a similar set of permutations for each partitioning—each would have $4$ elements. So if we just divide the total number of permutations by the number of permutations in each partitioning (i.e. $4$), we should get the number of partitionings!

Do you see why that is? It’s because that’s what division does!² Indeed $\frac{4!}{4} = 6$ which is the number of partitionings.³

How did we get $4$? Well, if we think of the number of ways to arrange elements in a particular partitioning, we get $2! 2!$. In general, it would be $k_1! k_2! \cdots k_r!$.

So the answer would be:

\[\binom{n}{k_1,\, k_2,\, \ldots,\, k_r} = \frac{n!}{k_1! k_2! \cdots k_r!}\]

And that’s the multinomial coefficient!

There are actually $6$ ways to do it if we care about the ID of the groups ($2$ for each of the partitionings we listed). In this blog post, we’ll just concern ourselves when the order of the groups doesn’t matter. ↩
Division tells you: if we divide $n$ things into groups of size $k$, how many groups are there?⁴ ↩
This actually gives us the number of partitionings if we consider that the groups are different (e.g. every group has a group ID). In this case, the groups are all the same, so we divide by an additional $2!$ (the number of ways to order two groups). ↩
Alternatively, it says if we divide $n$ things into $k$ groups, how many elements are in each group? ↩