# Symmetry Rule for Gaussian Binomial Coefficients

## 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 $0 \le k \le n$.

Consider the case $k \le \dfrac n 2$.

Then $k \le n - k$.

 $\ds \binom n {n - k}_q$ $=$ $\ds \prod_{j \mathop = 0}^{\paren {n - k} - 1} \frac {1 - q^{n - j} } {1 - q^{j + 1} }$ Definition of Gaussian Binomial Coefficient $\ds$ $=$ $\ds \paren {\frac {1 - q^{n - 0} } {1 - q^{0 + 1} } } \paren {\frac {1 - q^{n - 1} } {1 - q^{1 + 1} } } \paren {\frac {1 - q^{n - 2} } {1 - q^{2 + 1} } } \cdots \paren {\frac {1 - q^{n - \paren {\paren {n - k} - 1} } } {1 - q^{\paren {\paren {n - k} - 1} + 1} } }$ $\ds$ $=$ $\ds \paren {\frac {1 - q^{n - 0} } {1 - q^{0 + 1} } } \paren {\frac {1 - q^{n - 1} } {1 - q^{1 + 1} } } \paren {\frac {1 - q^{n - 2} } {1 - q^{2 + 1} } } \cdots \paren {\frac {1 - q^{n - \paren {k - 1} } } {1 - q^{\paren {k - 1} + 1} } } \paren {\frac {1 - q^{n - k} } {1 - q^{k + 1} } } \cdots \paren {\frac {1 - q^{k + 1} } {1 - q^{n - k} } }$ $\ds$ $=$ $\ds \paren {\frac {1 - q^{n - 0} } {1 - q^{0 + 1} } } \paren {\frac {1 - q^{n - 1} } {1 - q^{1 + 1} } } \paren {\frac {1 - q^{n - 2} } {1 - q^{2 + 1} } } \cdots \paren {\frac {1 - q^{n - \paren {k - 1} } } {1 - q^{\paren {k - 1} + 1} } }$ The tail cancels out $\ds$ $=$ $\ds \binom n k_q$ Definition of Gaussian Binomial Coefficient

The case $k \ge \dfrac n 2$ can be done by observing:

$n - k \le \dfrac n 2$

and hence by the above:

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

$\blacksquare$