Moving Top Index to Bottom in Binomial Coefficient

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Theorem

Let $m \in \Z, n \in \Z: n \ge 0$.

Then:

$\displaystyle \binom n m = \left({-1}\right)^{n-m} \binom {- \left({m + 1}\right)} {n - m}$

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \binom n m$	$=$	$\displaystyle \binom n {n - m}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Symmetry Rule for Binomial Coefficients
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \left({-1}\right)^{n-m} \binom {\left({n - m}\right) - n - 1} {n - m}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Negated Upper Index of Binomial Coefficient
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \left({-1}\right)^{n-m} \binom {- \left({m + 1}\right)} {n - m}$	$\displaystyle $	$\displaystyle $	$\displaystyle $

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \binom n m\)	\(=\)	\(\displaystyle \binom n {n - m}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Symmetry Rule for Binomial Coefficients
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({-1}\right)^{n-m} \binom {\left({n - m}\right) - n - 1} {n - m}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Negated Upper Index of Binomial Coefficient
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({-1}\right)^{n-m} \binom {- \left({m + 1}\right)} {n - m}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)