Symmetry Rule for Gaussian Binomial Coefficients

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Theorem

Let $q \in \R_{\ne 1}, n \in \Z_{>0}, k \in \Z$.

Then:

$\dbinom n k_q = \dbinom n {n - k}_q$

where $\dbinom n k_q$ is a Gaussian binomial coefficient.


Proof

If $k < 0$ then $n - k > n$.

Similarly, if $k > n$, then $n - k < 0$.


In both cases:

$\dbinom n k_q = \dbinom n {n - k}_q = 0$


Let $0 \le k \le n$.

\(\displaystyle \binom n {n - k}_q\) \(=\) \(\displaystyle \prod_{j \mathop = 0}^{\left({n - k}\right) - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }\) Definition of Gaussian Binomial Coefficient
\(\displaystyle \) \(=\) \(\displaystyle \left({\dfrac {1 - q^{n - 0} } {1 - q^{0 + 1} } }\right) \left({\dfrac {1 - q^{n - 1} } {1 - q^{1 + 1} } }\right) \left({\dfrac {1 - q^{n - 2} } {1 - q^{2 + 1} } }\right) \cdots \left({\dfrac {1 - q^{n - \left({\left({n - k}\right) - 1}\right)} } {1 - q^{\left({\left({n - k}\right) - 1}\right) + 1} } }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\dfrac {1 - q^{n - 0} } {1 - q^{0 + 1} } }\right) \left({\dfrac {1 - q^{n - 1} } {1 - q^{1 + 1} } }\right) \left({\dfrac {1 - q^{n - 2} } {1 - q^{2 + 1} } }\right) \cdots \left({\dfrac {1 - q^{k + 1} } {1 - q^{n - k} } }\right)\)
Then some magic happens ...
\(\displaystyle \) \(=\) \(\displaystyle \left({\dfrac {1 - q^{n - 0} } {1 - q^{0 + 1} } }\right) \left({\dfrac {1 - q^{n - 1} } {1 - q^{1 + 1} } }\right) \left({\dfrac {1 - q^{n - 2} } {1 - q^{2 + 1} } }\right) \cdots \left({\dfrac {1 - q^{n - \left({k - 1}\right)} } {1 - q^{\left({k - 1}\right) + 1} } }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \binom n k_q\)


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