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Are there any known examples of combinatorial objects that become easier to count by considering some kind of $q$-analog? It seems to me that it might be impossible for the problem of computing the $q$-analog directly to be (strictly) easier than enumerating the objects themselves, as we should be able to just replace $q$ everywhere with 1. However, I'd also be interested in any enumerative problems in which $q$-analogs give us some additional insight.

azimut
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LCL
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  • Isn't a large part of the point of $q$-analogs to 'give additional insight'? Keep in mind that you might not be able to directly plug $q=1$ into a $q$-formula and evaluate (e.g., even the simplest notion, the "$q$-number" $[n]=\frac{q^n-1}{q-1}$, either needs to be explicitly divided out or needs to have a smart limit taken), so I can imagine that there might be cases where it's easier to find the $q$ version and then take $\lim_{q\to 1}$ to derive a classical value rather than getting the classical value directly. – Steven Stadnicki Sep 16 '13 at 21:58
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    $q$-binomial coefficients tell how many are there $k$-dimensional linear subspaces of ${\mathbb F}_q^n$. Does this count? – sdcvvc Sep 16 '13 at 22:45

2 Answers2

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Sometimes the answer to a combinatorial problem is given by a ratio where both numerator and denominator can be computed as polynomials in $q$ but become zero for $q=1$.

One non-trivial example: the q-number of plane partitions that fit into $a\times b\times c$ box is given by the principal specialization of a Schur polynomial — i.e. it's a ratio of two Vandermonde determinants. This yields MacMahon formula.

(Slightly more generally, it's easier to proof q-hook-content formula first and get ordinary hook-content formula by plugging $q=1$.)

Grigory M
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It seems to me that it might be impossible for the problem of computing the $q$-analog directly to be (strictly) easier than enumerating the objects themselves, as we should be able to just replace $q$ everywhere with $1$.

It's not impossible, because the $q$ proof does not necessarily specialize to a proof at $q=1$. Maybe some objects in the proof collapse, maybe some arguments that don't only involve counting become incorrect, and for answers that are more complicated than setting $q=1$ in a polynomial, showing convergence and correctness of the answer as $q \to 1$ (or however else the solution at $1$ is obtained) can be more difficult than the $q$ argument.

For example, the $q$-gamma function is easier to define and handle than $\Gamma(x)$, and does not need any regularization, when $|q| \neq 1$, but justifying its relation to the gamma function and properties thereof, is more complicated.

zyx
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  • (Re: 'justifying its relation to the gamma function... is more complicated'): the definition of q-gamma is completely analogous to a definition of ordinary gamma-function (the lesser-known-but-easier-to-comprehend one). – Grigory M Jun 10 '14 at 21:23
  • The q-gamma is undefined at q=1, there are (1-q^a) factors everywhere in the numerator and denominator. Showing that the q=1 limit exists and is equal to the ordinary gamma function, and that properties of q-gamma become properties of ordinary gamma, is the "justifying [that] is more complicated" than working with the q-object by itself. – zyx Jun 10 '14 at 21:38