The Generalized Binomial Coefficient Formula

Recall from the Binomial Coefficients and The Binomial Theorem pages that we say $\binom{n}{k} = \frac{n!}{k!(n - k)!}$ is a binomial coefficient because for the binomial $(x + y)^n$ we have that $\binom{n}{k}$ is the coefficient attached to the simplified term $x^{n-k}y^k$ after the expansion of this binomial since for all $x, y \in \mathbb{R}$ and $n \in \{0, 1, 2, ... \}$ we saw that:

(1)

\begin{align} \quad (x + y)^n = \binom{n}{0} x^ny^0 + \binom{n}{1} x^{n-1}y^1 + ... + \binom{n}{n-1} x^1y^{n-1} + \binom{n}{n} x^0y^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \end{align}

We will now briefly touch upon the generalized binomial coefficient formula which still allows us to compute $\binom{n}{k}$ for $n, k \in \{0, 1, 2, ..., n \}$ and $0 \leq k \leq n$, however, it will allow $n$ to be any real number and for $k$ to be any integer. For $n \in \mathbb{R}$ and $k \in \mathbb{Z}$, the generalized binomial coefficient formula is:

(2)

\begin{align} \binom{n}{k} = \begin{Bmatrix} \frac{n \cdot (n-1) \cdot (n-2) \cdot ... \cdot (n-k+1)}{k \cdot (k-1) \cdot (k-2)...2 \cdot 1} = \prod_{j=1}^{k} \frac{n-j+1}{j} && \mathrm{if} \: k \geq 0\\ 0 && \mathrm{if} \: k < 0 \end{Bmatrix} \end{align}

Let's verify the formula above gives us the same value as $\binom{5}{3} = \frac{5!}{3!(5 - 3)!} = \frac{5!}{3! \cdot 2!} = 10$. We have that $n = 5$ and $k = 3$ and so with the formula we get:

(3)

\begin{align} \quad \prod_{j=1}^{3} \frac{5 - j + 1}{j} = \frac{5}{1} \cdot \frac{4}{2} \cdot \frac{3}{3} = 5 \cdot 2 \cdot 1 = 10 \end{align}