How do we get this equality?

\begin{eqnarray*} \binom{n+l-1}{l}q^l=\binom{-n}{l}(-q)^l \end{eqnarray*}

I haven't worked a lot with the binomial coefficient, so I don't really have a feel for these things... I really hope someone would help me with this, even though I haven't shown what I've tried, because I haven't been able to come up with much.

  • $\begingroup$ how did you get the coefficient of n negative $\endgroup$ – Nebo Alex Jan 11 '17 at 9:47

Why, that's simple. $$\binom{n+l-1}{l}={\overbrace{(n+l-1)(n+l-2)\dots n}^{\color{red}{l\;terms}}\over l!}$$ On the other hand, $$\binom{-n}{l}={\overbrace{(-n)(-n-1)\dots(-n-l+1)}^{\color{red}{l\;terms}}\over l!}={n(n+1)\dots(n+l-1)\over l!}\cdot(-1)^l$$

  • $\begingroup$ It is one of the basic binomial identity, called "Upper Negation" $\endgroup$ – G Cab Jan 11 '17 at 10:26
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    $\begingroup$ Oh, I had completely forgotten to write it out as a quotient with factorials. Thank you a lot for your help. $\endgroup$ – Sha Vuklia Jan 11 '17 at 10:32

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