This was posed as a challenging exercise by my professor. The exercise is to compute the dimension of the invariant subspace of $(\mathbb{C}^2)^{\otimes 2n}$ viewed as an $\mathfrak{sl}(2,\mathbb{C})$ representation.
For some background, $\mathfrak{sl}(2,\mathbb{C})$ can be viewed as traceless $2 \times 2$ matrices with complex entries. So one common basis we see is
$e = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, f = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.
The standard representation of $\mathfrak{sl}(2,\mathbb{C})$ on $\mathbb{C}^2$ is just by matrix multiplication. Let $v_1 = (1,0)$ and $v_0=(0,1)$. Observe that $ev_1 = fv_0 = 0$, $ev_0 = v_1,fv_1 = v_0$, and $hv_1 = v_1,hv_0 = -v_0$.
In general, if we have a Lie algebra $\mathfrak{g}$ acting on two representations $V,W$, then we can define a representation on their tensor product by a Leibniz rule: $a(v \otimes w) = (av) \otimes w + v \otimes (aw)$.
An invariant subspace of a representation $V$ is really all vectors that are annihilated by every element of $\mathfrak{g}$: $V^\mathfrak{g} := \{v \in V: av = 0, \forall a \in \mathfrak{g}\}$.
Okay, with that set up, some knowledge of $\mathfrak{sl}(2,\mathbb{C})$ tells us that the Lie algebra is completely reducible and all the irreducible complex representations are of the form $V(n) = \langle v_n, v_{n-2}, v_{n-4},...,v_{2-n},v_{-n} \rangle$, a $n+1$ dimensional vector space. The subscripts here represent the weight which really means that they are eigenvectors of $h$ with their eigenvalue being the subscript. So $h v_{n-6} = (n-6) v_{n-6}$. I didn't write it this way above but it's mostly due to bookkeeping.
Moreover, $e$ adds $2$ to the weight of vectors while $f$ subtracts $2$ from their weight. $h$ doesn't shift the weight. Anyways, the relevance of this is that the invariant subspace must lie within the weight $0$ subspace because otherwise, $h$ cannot annihilate. Also, under tensor products, weights add.
For example, $\mathbb{C}^2 \otimes \mathbb{C}^2$ has basis: $v_1 \otimes v_1, v_1 \otimes v_0, v_0 \otimes v_1, v_0 \otimes v_0$. The first one has weight $2$, the middle two have weight $0$, the last one has weight $-2$. We can show that $e(v_1 \otimes v_0 - v_0 \otimes v_1) = 0$ and $f$ and $h$ also annihilate this vector. Well, $h$ acts by multiplication by $0$ on the weight $0$ subspace so we don't really need to check $h$.
Note also that if we take $(\mathbb{C}^2)^{\otimes n}$ where $n$ is odd, then there are no weight $0$ vectors other than the zero vector. In this case, there cannot be any nontrivial invariant subspace. That's why we take even number of copies.
Now the bookkeeping: I chose $v_1,v_0$ to denote the basis vectors of $\mathbb{C}^2$ because then we can denote the basis elements of $W_{2n} := (\mathbb{C}^2)^{\otimes 2n}$ by strings of zeros and ones. So $\dim W_{2n} = 2^{2n}$. Among these, the weight $0$ strings are precisely those with half being zeros, the other half being ones. So the weight zero subspace has dimension ${2n \choose n}$. The invariant subspace has to have dimension smaller than this. I could have chosen $+,-$ instead but it might get confusing to read things like $(+-)-(-+)$.
Also, as an example, in $W_6$, $e(101001) = (111001) + (101101) + (101011)$. Basically, $e$ flips $0$ to $1$ like in computers and but kills $1$; that's why we see only three terms instead of six which we expect from Leibniz's rule. $f$ does the opposite. I suspect that we really only need to study $e$ but I'm not sure.
Anyways, with that lengthy prologue (if you read till here, thank you), I admit that I don't see any combinatorial patterns. I thought I had found something when $n=2$ but it fell apart in $n=3$. For $n=2$, I think the invariant subspace has dimension $2$ and is spanned by $(1100)-(1010)-(0101)+(0011)$ and $(1001)-(1010)-(0101)+(0110)$.