Is this $\binom{n}{p}$ for $p>n$ make a sense in mathematics or it is $0$ by convention?

Question

It is well known that gamma function is not defined at negative integers , but my question is to know how i take the value of $\binom{n}{p}$ for $p>n$ then is this make a sense or it is $0$ by convention ?

Answer 1

Define $\binom{n}{p}$ as the number of subsets of $\{1, \ldots, n\}$ having exactly $p$ elements.

Then it makes mathematical sense to say that $\binom{n}{p}=0$ if $p>n$. Of course, if you choose this definition then you have to prove that $\binom{n}{p} = \frac{n!}{p!(n-p)!}$ for all $p, n \in \mathbb N$ such that $p \leq n$.

Answer 2

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).

Answer 3

If you directly define $\binom{n}{p}$ as $\frac{\Gamma(n+1)}{\Gamma(p+1)\Gamma(n-p+1)}$ and recall that $\Gamma(x)$ ha simple poles at $0,-1,-2,\ldots$, it is no wonder that $\binom{n}{p}=0$ for $p,n\in\mathbb{N}$ and $p>n$. It is usually introduced by convention, but it is also the unique convention which agrees with the analytic continuation of the binomial coefficients through the $\Gamma$ function.