"C choose k" where C is topological space One day I read a generating function in a paper.  For any "sufficietly nice topological space", $C$:
$$  \sum_{l \geq 0 } q^{2l}\chi(\mathrm{Sym}^l[C]) = (1 - q^2)^{-\chi(C)} = \sum_{l \geq 0} \binom{l + \chi(C)-1}{l} q^{2l} $$
First of all, I'm not sure what "sufficiently nice" means here.  I'm guessing any CW complex will do.  I wonder what's an example of a space where this formula doesn't work. 

This formula suggests 
$$ \chi(\mathrm{Sym}^l[C]) = \binom{l -1+ \chi(C)}{l} = \chi \binom{l-1 + C  }{l} $$
where the right hand side is some "categorification".     Removing the $\chi$'s, is there some sense in which 
$$ \mathrm{Sym}^l[C] = \binom{l -1+ C  }{l} $$
is rigorous?
 A: The papers I was thinking of are actually by Propp:

*

*Euler measure as generalized cardinality, and

*Exponentiation and Euler measure
(although I think these papers are outdated and more is known these days.)  One has to restrict to a certain class of particularly nice spaces and modify the definition of Euler characteristic.
A: I presume you are asking is whether one can make sense of $\binom{X}{k}$ as a space in such a way that it relates to the formula you are asking about.
The answer is yes. 
First let $l = 1$. For a space $X$ define $\binom{X}{k}$ to be the configuration space of subsets of X
having cardinality $k$.  Then $\binom{X}{1} = X$. In this case $\text{Symm}^1(X) = X$ 
and we have agreement
$$
\binom{X}{1} = X = \text{Symm}^1(X) .
$$
Now consider the case $l=2$, and let {1} be the one element set
Then as sets there is an evident bijection
$$
\binom{X \amalg \text{{1}} }{2} = \binom{X}{2} \amalg (X\times \text{{1}}) 
$$
As a set, the right side is the same thing as $\text{Symm}^2(X) = X\times_{Z_2} X =$ the orbits of the cyclic group of order two acting on $X\times X$ by permutation. To see this, note that $X\times X$ has two kinds of isotropy: one coming from 
the diagonal copy of $X$ (with trivial action) and the other being it's complement 
which is $X\times X - X$ with free action having quotient $\binom{X}{2}$. 
With respect to this identification, your formula makes sense. However, these spaces
have different topology when $X$ isn't discrete.
A similar observation works in the $l >2$ case.
