Reflection Rule for Gaussian Binomial Coefficients

Theorem

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

Then:

$\dbinom n k_q = q^{k \paren {n - k} } \dbinom n k_{q^{-1} }$

$\ds \dbinom n k_{q^{-1} }$

$=$

$\ds \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - \paren {q^{-1} }^{n - j} } {1 - \paren {q^{-1} }^{j + 1} }$

$\ds $

$=$

$\ds \prod_{j \mathop = 0}^{k - 1} \dfrac {\frac {q^{n - j} - 1} {q^{n - j} } } {\frac {q^{j + 1} - 1} {q^{j + 1} } }$

$\ds $

$=$

$\ds \prod_{j \mathop = 0}^{k - 1} \dfrac {q^{j + 1} } {q^{n - j} } \dfrac {q^{n - j} - 1} {q^{j + 1} - 1}$

$\ds $

$=$

$\ds \prod_{j \mathop = 0}^{k - 1} q^{1 - n + 2 j} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }$

$\ds $

$=$

$\ds \paren {q^{1 - n} }^k \prod_{j \mathop = 0}^{k - 1} q^{2 j} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }$

$\ds $

$=$

$\ds q^{k - n k} q^{k \paren {k - 1} } \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }$

Closed Form for Triangular Numbers: $\ds \sum_{j \mathop = 0}^{k - 1} 2 j = 2 \sum_{j \mathop = 1}^{k - 1} j = 2 \dfrac {\paren {k - 1} k} 2$

$\ds $

$=$

$\ds q^{k^2 - n k} \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }$

$\ds $

$=$

$\ds \dfrac 1 {q^{k \paren {n - k} } } \dbinom n k_q$

$\blacksquare$