I have already looked at this proof here

Proving that $n \choose k$ is an integer

However I don't understand how I can use the Pascal identity for binommial coefficients if $n$ is a negative number.

I have had the idea to prove a Connection between the negative and the positive Counterpart.


Maybe $\binom{n}{k}=-x\binom{-n}{k}$, where $x\in\mathbb{Z}$

I wrote down


What could be the $x$ ?

Thanks for reading.

  • $\begingroup$ $\frac{1}{k!}n(n-1)…(n-k+1)=\binom{n}{k}$ is the Definition for General binomial coefficients $\endgroup$ – New2Math Mar 18 at 14:20

For $n>0$ we have \begin{align} \binom{-n}k &=\frac{-n(-n-1)\cdots(-n-k+1)}{k!}\\ &=(-1)^k\frac{n(n+1)\cdots(n+k-1)}{k!}\\ &=(-1)^k\frac{(n+k-1)!}{k!(n-1)!}\\ &=(-1)^k\binom{n+k-1}{k} \end{align} hence it is an integer.

  • 2
    $\begingroup$ Can the downvoter please say what is wrong with this answer? $n>1$ for the case $n=0$ one can make a seperate Argument, Therefore. $n-1>0$ and a integer and therefore $n-1+k$ is a natural number. And then with the already known result the Statement is true $\endgroup$ – New2Math Mar 18 at 14:30
  • $\begingroup$ You can't write like this! $\endgroup$ – LAGRIDA Mar 18 at 14:32
  • $\begingroup$ The generale formula of factorial : $\Gamma(n+1) = n!$, for $n$ negative $n! = \pm \infty$ $\endgroup$ – LAGRIDA Mar 18 at 14:34
  • $\begingroup$ @LAGRIDA: in my answer, there are no $n!$ with $n<0$. $\endgroup$ – Fabio Lucchini Mar 18 at 14:37
  • $\begingroup$ but the definition : $n! = n (n-1) \cdots 2$ true iff n > 1 $\endgroup$ – LAGRIDA Mar 18 at 14:39

Alternative method:

Note that

$\binom{n}{k} = \binom{n-1}{k-1}+ \binom{n-1}{k} \space \forall n \in \mathbb{Z} \space \forall k \in \mathbb{N} \\ \Rightarrow \binom{n-1}{k} = \binom{n}{k}- \binom{n-1}{k-1} \space \forall n \in \mathbb{Z} \space \forall k \in \mathbb{N}$

Together with $\binom{n}{0}=1 \space \forall n \in \mathbb{Z}$ you can use this to prove by induction that $\binom{n}{k} \space 0>n\in\mathbb{Z}, k\in \mathbb{N}$ is the difference of two integers and so is always an integer.


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