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*. 


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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". 

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 Removing the $\chi$'s, is there some sense in which 
$$ \mathrm{Sym}^l[C] = \binom{l -1+ C }{l} $$
is rigorous?



"C choose k" where C is topological space
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.


