We all have known that $\binom{n} {k} = \dfrac{n!}{k! (n-k)!}$ which is a combination standard formula for $n \geq k$. But, recently I myself encounter this binomial notation but for $n \leq k$. Now, how do you express this in factorial form? Doesn't it leads to undetermined form because no negative integers is defined in factorial forms. Also, I ever see that $n$ here is rational number, not an integer number, making me confuse how to calculate it later. I am thinking that there should be new definition which expands this, yet I don't learn it. Please, could you explain this briefly?
2026-03-25 12:45:47.1774442747
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Binomial Notation Expansion
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A common definition of the binomial coefficient with $\alpha\in\mathbb{C}$ and integer values $p$ is \begin{align*} \binom{\alpha}{p}= \begin{cases} \frac{\alpha(\alpha-1)\cdots(\alpha-p+1)}{p!}&p\geq 0\\ 0&p<0 \end{cases} \end{align*}
From this we conclude $\binom{n}{p}=0$ if $p>n \ \ (n,p\in\mathbb{N})$.
Hint: The chapter 5 Binomial coefficients by R.L. Graham, D.E. Knuth and O. Patashnik provides a thorough introduction. The formula above is stated as (5.1).
As long as $k$ is a positive integer, we can use the alternate form $$ \binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} $$ This also works for $n\notin \Bbb N$.
For other uses, we must use the $\Gamma$ function, defined as $$ \Gamma(x) = \int_0^\infty t^{x-1}e^{-t}\,dt $$ This function has the property that $\Gamma(n+1) = n!$ for integers $n>0$. Thus we can define $$ \binom{n}{k} = \frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)} $$ The $\Gamma$ function is defined for all real numbers apart from $0$ and the negative integers. So as long as $k-n$ is not a positive integer this definition works. (Also, we need $n$ and $k$ to not be negative integers, of course.)
In the cases where $k-n$ is a positive integer, it can be argued that $\binom nk = 0$ makes sense. One reason is that the numerator is finite while the denominator goes to $\pm\infty$, so that's the limit we get. The second reason is that this generalizes the convention that for the regular binomial coefficients (with $n, k\in \Bbb N$) we have $\binom nk = 0$ if $k>n$.