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In Stanley, EC2, we have the following statement:

screenshot from corollary 7.23.9

I think there is a typo in the first sum after "generating function", and that $[n]_q!$ should be replaced by $(1-q)(1-q^2)\dotsb (1-q^n)$, and same for $[n]_t!$.

The problem with the proof seems to be the line

line of proof

which is incorrect(?) as the denominator on the right hand side should be not the q/t-factorials, but the above product, according to the lemma which is referenced there.

This issue (if my suspicions hold), is not listed in the Errata..

I have also checked this in Mathematica, so I am fairly confident this is a typo.

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    $\begingroup$ Looks like it. It's easy to mix up $[n]_q!$ and $(1-q)(1-q^2)\cdots(1-q^n)$ (which of course differ only by a factor of $(1-q)^n$): that's a mistake that I make all the time when writing these kind of generating functions. $\endgroup$
    – Sam Hopkins
    Commented Feb 17, 2021 at 14:33
  • $\begingroup$ @SamHopkins yes, exactly my thought... $\endgroup$
    – Per Alexandersson
    Commented Feb 17, 2021 at 15:11
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    $\begingroup$ Thanks, I will fix this. $\endgroup$
    – Richard Stanley
    Commented Feb 17, 2021 at 19:42
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    $\begingroup$ On second thought, just before Corollary 7.23.9 I define $[n]!_q=(1-q)(1-q^2)\cdots (1-q^n)$, and similarly for $[n]!_t$. My notation for $[n]!_q/(1-q)^n$ is $(n)!$ (in boldface---MO displays all math in boldface, which doesn't work so well for my notation). $\endgroup$
    – Richard Stanley
    Commented Feb 18, 2021 at 2:51
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    $\begingroup$ @RichardStanley Ah, I see! My mistake for not looking up the convention! $\endgroup$
    – Per Alexandersson
    Commented Feb 18, 2021 at 7:41

1 Answer 1

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(In order that this question appears as answered I'm converting Richard's comment into an answer.)

The (somewhat nonstandard) convention in the text, stated just before the corollary, is that $[n]!_q = (1-q)(1-q^2)\cdots(1-q^n)$. For the quantity $[n]!_q/(1-q)^n$, the notation $\mathbf{(n)!}$ is used instead.

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