Multinomial coefficients are often denoted as \( \left( n;~ k_1, \dots, k_m \right) \) where \( n = \sum_{i=1}^m k_i \). Another way to express a multinomial coefficient is using the \( {\rm choose} \) notation commonly employed in combinatorics for binomial coefficients. Given \( n \geq \sum_{i=1}^{m-1} k_i \), we say that the following are equivalent:

$$ \left( n;~ k_1, \dots, k_m \right) ~\mbox{where}~ k_m = n - \sum_{i=1}^{m-1} k_i \equiv {n \choose k_1, \dots, k_{m-1}} $$

For convenience, I'm going to use the first notation.

### Theorem

I'm not sure if this has ever been stated before, but here's an observation that I made a couple months ago:

$$ \left( n;~ k_1, \dots, k_m \right) = \left( n-k_m;~ k_1, \dots, k_{m-1} \right) \cdot \frac{1}{k_m !} \prod_{i=1}^{k_m} n-i+1 $$

That is, there exists a linear, tail recursive definition for calculating multinomial coefficients.

### Proof of Theorem

Without loss of generality, assume that \( n = \sum_{i=1}^{m} k_i \). According to the equivalence noted above as well as the Multinomial Theorem, we have:

$$ \left( n;~ k_1, \dots, k_m \right) = {n \choose k_1, \dots, k_m} = \frac{ n! }{ k_1 ! \cdot \cdots \cdot k_m ! } $$

(1) If we pull out the term \( k_m \), we get:

$$ \left( n;~ k_1, \dots, k_m \right) = \frac{1}{k_m !} \cdot \frac{ n! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! } $$

(2) Now, let's find some \( \zeta \) such that:

$$ \frac{ (n-k_m)! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! } \cdot \zeta = \frac{ n! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! } $$

(3) If we isolate \( \zeta \) on the left-hand side, we observe the following:

$$ \zeta = \frac{ n! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! } \cdot \frac{ k_1 ! \cdot \cdots \cdot k_{m-1} ! }{ (n-k_m)! } = \frac{ n! }{ (n-k_m)! }$$

(4) We know that certain terms cancel in this fraction. It is easy to see that

$$ \zeta = \frac{ n! }{ (n-k_m)! } = (n - k_m + 1) \cdot (n - k_m + 2) \cdot \cdots \cdot n = \prod_{i=1}^{k_m} n - k_m + i $$

(5) Since the product in \( \zeta \) is from \( 1 \) to \( k_m \), we can be further simplify the expression to

$$ \zeta = \prod_{i=1}^{k_m} n - k_m + i = \prod_{i=1}^{k_m} n-i+1 $$

(6) Replacing \( \zeta \) in (2) gives us the following:

$$ \frac{ (n-k_m)! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! } \cdot \prod_{i=1}^{k_m} n-i+1 = \frac{ n! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! } $$

(7) Now we use the result of (6) with (1) to get:

$$ \left( n;~ k_1, \dots, k_m \right) = \frac{1}{k_m !} \cdot \frac{ (n-k_m)! }{ k_1 ! \cdot \cdots \cdot k_{m-1} ! } \cdot \prod_{i=1}^{k_m} n-i+1 $$

(8) Since the second fraction in (7) can be expressed as a multinomial, we have:

$$ \left( n;~ k_1, \dots, k_m \right) = \left( n-k_m;~ k_1, \dots, k_{m-1} \right) \cdot \frac{1}{k_m !} \prod_{i=1}^{k_m} n-i+1 $$

Therefore, the theorem is correct.

### Some Notes

This result that we observed in (7) directly relates to the ratios between the coefficients on the face of a multidimensional Pascal triangle.

### Corollary

It is easy to see that the following follows from the theorem:

$$ \left( n;~ k_1, \dots, k_m \right) = \left( n-k_m;~ k_1, \dots, k_{m-1} \right) \cdot \prod_{i=1}^{k_m} \frac{n-i+1}{i} $$

### Folding with scala

In Scala, the expression in the corollary can be expressed concisely using the foldLeft operator.This operator applies a binary operator to a start value and all elements of a list, going left to right. If we let our list be the numbers \( 1, \dots, k_m \), then we can foldLeft with the following binary operator:

$$ \left(a,b\right) \Rightarrow \frac{a \left(k_m-b+1\right)}{b} $$

For more information on the general fold operator (with which you can pretty much do anything with), see "Introduction to Metamathematics" by Kleene (affiliate link).

### The Code

Here is my recursive implementation of the recurrence relation mentioned in the corollary above:

I've omitted a an iterative implementation of the corollary that I've written for the sake of brevity. If anyone expresses some interest, I may include here or in a future blog post.