Let $A=(a_{i_1\dots i_k})_{i_1,\dots,i_k=1}^n$ be a higher order cubic tensor or hypermatrix. The following two facts are well-known and are easy to prove:
${(\bf 1) }$ The dimension of the subspace of the completely symmetric tensors (that is, $a_{i_1\dots i_k}$ is the same for any permutation of $i_1,\dots,i_k$ ) is equal to the number of $k$ combinations with repetition = $\frac{(n+k-1)!}{k!(n-1)!}$.
${(\bf 2) }$ The dimension of the subspace of the completely skew-symmetric tensors (that is, $a_{i_1\dots i_k}$ only changes sign for odd permutations of $i_1,\dots,i_k$ ) is equal to the number of $k$ combinations = $\frac{n!}{k!(n-k)!}$.
${\bf Question:}$ My question is, how to compute the dimensions in the cases between (1) and (3). Namely, what is the dimension of the subspace of tensors obtained by making some partial symmetrizations of $A$ and then making partial skew-symmetrizations of the result.
I do not know the answer even in the third-order case. For example, if we start from the subspace of the $n\times n\times n$ tensors and make the partial symmetrization $a_{i_1i_2i_3}\rightarrow (a_{i_1i_2i_3} + a_{i_2i_1i_3})/2$, then we obviously, obtain the subspace of dimension $\frac{n(n+1)}{2}n$. What is the dimension of the new subspace obtained by taking one more skew-symmetrization $a_{i_1i_2i_3}\rightarrow (a_{i_1i_2i_3} - a_{i_1i_3i_2})/2$? (that is, first we symmetrize the frontal slices, and then we skew-symmetrize the horizontal clises).
Is there a formula that gives the dimension of the image of the space of the $k$-th order tensors in terms of partial symmetrizations and skew-symmetrizations?