What is the number of direct sum decompositions of an n-dimensional vector space into k subspaces? The answer to the above question is given in Sloane's OEIS A270880.
I am trying to read the journal article given below.  The first formula in section 1.4 on page 4 is supposed to give these numbers.  I think the formula is wrong.  Does anyone agree/disagree?  
Kent E. Morrison, Integer Sequences and Matrices Over Finite Fields, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.1. 
 A: One way to check if the formula is correct is to evaluate $\displaystyle \lim_{q \to 1} {n \brace k}_{q}$ and check if it is equal to $\displaystyle {n \brace k}.$
Using the formulae given in the paper you linked to, this is true as seen from the computation below: (the summation is over all $k-$compositions of $n,$ that is over all $k-$tuples $(n_1, \ldots , n_k)$ such that $n_1 + \ldots n_k = n$ and $n_i \geq 1$ for $i=1, \ldots ,k$) 
$\begin{aligned}
\displaystyle \lim_{q \to 1} {n \brace k}_{q}&=\dfrac{1}{k!}\displaystyle \lim_{q \to 1}\sum_{n_1 +\ldots n_k = n}\dfrac{\gamma_n}{\gamma_{n_{1}}\ldots\gamma_{n_{k}}}\\&=\dfrac{1}{k!}\displaystyle \lim_{q \to 1}\sum_{n_1 +\ldots n_k = n}\dfrac{(q-1)^{n}q^{\binom{n}{2}}[n]_{q}!}{(q-1)^{n_1}q^{\binom{n_1}{2}}[n_1]_{q}!\ldots(q-1)^{n_k}q^{\binom{n_k}{2}}[n_{k}]_{q}!}\\&=\dfrac{1}{k!}\displaystyle \lim_{q \to 1}\sum_{n_1 +\ldots n_k = n}\dfrac{(q-1)^{n}q^{\binom{n}{2}}[n]_{q}!}{(q-1)^{n_1 + \ldots + n_k}(q)^{\binom{n_1}{2}+\ldots\binom{n_k}{2}}[n_1]_{q}!\ldots[n_k]_{q}!}\\&=\dfrac{1}{k!}\sum_{n_1 +\ldots n_k = n}\displaystyle \lim_{q \to 1}\dfrac{q^{\binom{n}{2}}[n]_{q}!}{(q)^{\binom{n_1}{2}+\ldots\binom{n_k}{2}}[n_1]_{q}!\ldots[n_k]_{q}!}\\&=\dfrac{1}{k!}\sum_{n_1 +\ldots n_k = n}\displaystyle \dfrac{n!}{n_1!\ldots n_k!}\\&={n\brace k}\end{aligned}
$
In the first step, I used the first formula mentioned for $\displaystyle{n \brace k}_{q}.$
In the second step I used the second formula for $\gamma_{n}$
In the fourth step, I brought the limit inside. In the fifth step I used the fact that $\displaystyle \lim_{q \to 1}[n]_{q}! = n!$ and in the final step I used an equivalent expression of $\displaystyle{n \brace k}$ which can be obtained by by double-counting the number of surjective functions from an $n-$set of distinct objects to a $k-$set of distinct objects using multinomial coefficients via one way and the Stirling number of the second kind via the other way.
Hence the formula is correct.
A: The formula given in the article is correct.  Note that the summation is over all integer COMPOSITIONS of n, not partitions.
