\[\genfrac{[}{]}{0.0pt}{}{n}{m}_{q}=\frac{\left(q;q\right)_{n}}{\left(q;q\right)%
_{m}\left(q;q\right)_{n-m}}\\
=\frac{\left(q^{-n};q\right)_{m}(-1)^{m}q^{nm-\genfrac{(}{)}{0.0pt}{}{m}{2}}}{%
\left(q;q\right)_{m}},\]