The independent components of a tensor whose symmetry type is given by a given Young diagram each correspond to a semistandard tableau with that same shape. And the number of semistandard tableaux for a given Young diagram can be found via a hook length formula:
$$
N = \prod_{(i,j) \in \lambda} \frac{ k - i + j}{h_\lambda(i,j)}
$$
where (using the English conventions for Young diagrams)
- $k$ is the range of entries allowed in the tableau (in your case, the dimension of the underlying vector space)
- $i$ and $j$ are the index of the row and column, respectively (numbered starting from the top left, beginning with 1)
- $h_\lambda(i,j)$ is the hook length associated with the cell $(i,j)$; it is found by counting all the cells that are directly below the cell $(i,j)$ and all the cells that are directly to the right of $(i,j)$, and adding 1 for the cell itself. In other words, if you draw a line that comes from below the diagram, hits $(i,j)$, and then "hooks" to the right and exits the diagram, $h_\lambda(i,j)$ is the number of cells you passed through.
So, for a general $k$ and a Young diagram consisting of three cells in a single row, this becomes
$$
N = \underbrace{\left( \frac{k - 1 + 1}{3} \right)}_{(i,j) = (1,1)} \underbrace{\left( \frac{k - 1 + 2}{2} \right)}_{(i,j) = (1,2)} \underbrace{\left( \frac{k - 1 + 3}{1} \right)}_{(i,j) = (1,3)} = \frac{k(k+1)(k+2)}{6}
$$
which gives $N = 20$ when $k = 4$, as you found.
As another example, for the mixed-symmetry Young diagram with three cells, we have
$$
N = \underbrace{\left( \frac{k - 1 + 1}{3} \right)}_{(i,j) = (1,1)} \underbrace{\left( \frac{k - 1 + 2}{1} \right)}_{(i,j) = (1,2)} \underbrace{\left( \frac{k - 2 + 1}{1} \right)}_{(i,j) = (2,1)} = \frac{k(k+1)(k-1)}{3}.
$$
(This also happens to yield $N = 20$ when $k = 4$, but this is just a coincidence; for general $k$, the dimensionality of this representation is different.)
