Why should be accept this ${\displaystyle {n \choose k}}=0 $ for $k> n$?

Question

I'm confused how ${\displaystyle {n \choose k}}=0 $ for $k> n$ however the ratio :${\displaystyle\frac{n!}{(n-k)!k!}}$ is not defined for $k>n$ ?

My question here is: Why should be accept this $${\displaystyle {n \choose k}}=\frac{n!}{(n-k)!k!}=0 , for , k> n$$ ?

Note: I don't accept this if it were by convention because it's not defined !!!!

Thank you for any help

Answer 1

It’s the natural interpretation when $\binom{n}k$ is given its combinatorial interpretation as the number of $k$-element subsets of an $n$-element set. Moreover, outside very elementary treatments the binomial coefficient isn’t actually defined as

$$\binom{n}k=\frac{n!}{k!(n-k)!}\;;$$

a common and much more general definition is

$$\binom{x}k=\frac{x^{\underline k}}{k!}\;,$$

where $x^{\underline k}=x(x-1)(x-2)\ldots(x-k+1)=\prod_{i=0}^{k-1}(x-i)$. This definition is actually valid for all real $x$ and all non-negative integers $k$, and when $n$ is a non-negative integer less than $k$ we have $n^{\underline k}=0$, since one of the factors is $n-n$.

Answer 2

One of the possible definitions of $\dbinom nk$ is that it's the coefficient of $x^k$ when you expand $(1+x)^n$ as the product of $n$ factors (whence the name of binomial coefficient).

Now in this expansion you have no $x^k$ term if $k>n$ or if $k<0$.

Note the formula $\;\dbinom nk=\dfrac{n!}{k!\,(n-k)!}$ is only a theorem, valid for $0\le k\le n$.