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We all know that $\sum_{i=0}^{n}{n \choose i}=2^{n}$. Is there a similar result regarding the q-binomial coefficients? (a.k.a Gaussian binomial coefficients) - $\sum_{i=0}^{n}{n \choose i}_{q}=?$

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I think that you need to define your normalizations before stating the question. What is your definition of $[n]_q?$ – Scott Carter Mar 1 '12 at 7:54

4 Answers 4

The identity $\prod_{i=0}^{n-1} (1+xq^i) = \sum_{k=0}^n x^k q^{{k\choose 2}}{n\choose k}_q$ is the $q$-binomial theorem. A combinatorial proof based on integer partitions is mentioned on page 68 of Enumerative Combinatorics, vol. 1, 2nd ed. There is also given a combinatorial proof based on finite fields. For the online version at, see pages 74-75.

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There are many possibilities, e.g. $\sum_{i=0}^{n}q^i{n \choose i}_{q^2}=(1+q)(1+q^2)...(1+q^n)$


$\sum_{i=0}^{n}q^{i(i+1)/2}{n \choose i}_{q }=(1+q)(1+q^2)...(1+q^n).$

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That's interesting. Do you know a bijective proof of these identities, based on the fact that the Gaussian binomial coefficient is the number of subspaces of dimension $i$ in $(\mathbb{Z}/q)^n$? – Neil Strickland Mar 1 '12 at 10:38
I know no direct bijective proofs of this sort. The second identity can be deduced from a polynomial identity which has a bijective proof in the setting of finite vector spaces, cf. Goldman - Rota, Finite vector spaces and Eulerian generating functions, Studies in Appl. Math. XLIX (3), 1970. – Johann Cigler Mar 1 '12 at 12:36

For Gaussian binomial coefficients we have $$ \sum_{k = 0}^n \binom nk_q = \sum_{m = 0}^\infty a_m q^m, $$ where $$ a_m = \sum_{\lambda\vdash m} \#\{k\in \mathbf Z_{\geq 0}\mid \lambda_1\leq n-k, \lambda'_1\leq k\}. $$ The notation $\lambda\vdash m$ signifies that $\lambda$ is an integer partition of $m$. Also $\lambda_1$ is the first (largest) part of $\lambda$ and $\lambda'_1$ is the number of positive parts in $\lambda$.

This follows from the following well-known fact about Gaussian binomial coefficients: $$ \binom nk_q = \sum_\lambda q^{|\lambda|}, $$ the sum being over all partitions $\lambda$ where $\lambda_1\leq n-k$ and $\lambda'_1\leq k$ (see, for example Stanley's Enumerative Combinatorics, vol. 1, or Eq. (4) in my expository article titled Counting subspaces in a Finite Vector Space).

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One more version - analog of $\sum_{i=0}^n(-1)^i\binom ni=0$: $$ \sum_{i=0}^n(-1)^i\binom ni_q=\begin{cases}0,&n=2k-1\\ \prod_{j=1}^k(1-q^{2j-1}),&n=2k\end{cases} $$

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