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Show that ${-n \choose i} = (-1)^i{n+i-1 \choose i} $. This is a homework exercise I have to make and I just cant get started on it. The problem lies with the $-n$. Using the definition I get: $${-n \choose i} = \frac{ (-n)!}{i!(-n-i)!}$$ But how do I calculate $(-n)!$ in this expression? I know the Gamma function could be used but I am sure that is not what is required from us. Does anyone know how this works? Maybe it's just a question of definitions? Thanks in advance!

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    $\begingroup$ If that is the only definition you have for $\binom{-n}{i}$, yeah, you aren't going to be able to solve this question. You probably need a different definition of $\binom{x}{i}$ to deal with the cases where $x<0$ of $x$ is not an integer... $\endgroup$ Commented May 6, 2013 at 13:07

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When dealing with upper numbers that are not non-negative integers you should use the definition

$$\binom{x}k=\frac{x^{\underline k}}{k!}=\frac{x(x-1)(x-2)\dots(x-k+1)}{k!}\;,\tag{1}$$

where $x^{\underline k}$ is the falling factorial. It’s easy to check that when $x\in\Bbb N$, this agrees with the definition that you’re using. And it’s pretty easy to get the desired result from $(1)$; if you have any trouble with it, just leave a comment.

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  • $\begingroup$ Ah ok great, so just to be clear ${-4 \choose 3}=\frac{-4\cdot -5 \cdot -6 }{3!}$ as an example? $\endgroup$
    – Slugger
    Commented May 6, 2013 at 13:12
  • $\begingroup$ @Teun: Yes, that’s right. $\endgroup$ Commented May 6, 2013 at 13:13
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Yes, question of definitions:

For any real $\alpha$ and nonnegative integer $i$, we can define $${\alpha\choose i}:=\frac{\alpha\cdot(\alpha-1)\cdot \ldots\cdot (\alpha-i+1)}{i!} \,.$$

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  • $\begingroup$ Thanks a lot for your answer, I knew it wasn't anything too difficult! $\endgroup$
    – Slugger
    Commented May 6, 2013 at 13:14
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Note that $${n+i-1 \choose i }=\dfrac{(n+i-1)!}{i!(n-1)!}=\dfrac{(n+i-1)(n+i-2)\dots(n)}{i!} $$ and $$ \frac{(n+i-1)\cdots(n)}{i!} = (-1)^i \frac{(-n)(-n-1)(-n-2)\cdots(-n-i+1)}{i!} = (-1)^i{-n \choose i}$$

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  • $\begingroup$ I feel confused on $(-1)^i$. I can see that when i is odd, then it is the same. But when i is even? $\endgroup$
    – Mariana
    Commented Apr 2, 2021 at 15:58

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