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
$$
\binom{X \amalg \text{{1}} }{2} = \binom{X}{2} \amalg (X\times \text{{1}})
$$
which is a homeomorphism where we give the right side the topology induced from the left side. With this topology, 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 $\binom{X}{2}$.
A similar observation works in the $l >2$ case.