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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) - $\sum_{i=0}^{n}{n \choose i}_{q}=?$

flag	asked Mar 1 2012 at 7:47 Netanel 31●1

I think that you need to define your normalizations before stating the question. What is your definition of $[n]_q?$ – Scott Carter Mar 1 2012 at 7:54

Richard Stanley · Answer 1 · 2012-03-01 16:34:07Z UTC

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 http://math.mit.edu/~rstan/ec/ec1.pdf, see pages 74-75.

Johann Cigler · Answer 2 · 2012-03-01 09:55:33Z UTC

2

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

or

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

edited Mar 1 2012 at 9:55

answered Mar 1 2012 at 9:44

Johann Cigler
1,500●4●14

	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 2012 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 2012 at 12:36

Sum of Gaussian binomial coefficients.

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