Consider the irreducible representation of $\mathfrak{su}$(n) given by the Young tableu
The dimension of the representation is easily obtained by the usual rules, and it is $d=n(n-1)/2$. One can also directly guess $d$ by noting that the antisymmetric tensor product of two fundamental representations (with dimension $n$) has a base of $\begin{pmatrix} n \\ 2\end{pmatrix}=n(n-1)/2$ elements.
For concreteness, let me fix in the following $n=4$ and consider the totally symmetric product of two representations as above
where $\cdot$ is the trivial 1-dimensional representation ($n=4$ here)
With the rules for computing the dimension of the irreducible representations, the dimension of $\Gamma_1$ is $d(\Gamma_1) = n^2(n^2-1)/12 = 20$.
Alternatively, given the dimension $d=6$ of the representation in the first figure, I could have thought that the symmetric tensor product of two such representation has the dimension $\begin{pmatrix}d+2-1\\ 2\end{pmatrix}=\begin{pmatrix}7\\2\end{pmatrix}=21$.
What is wrong in the second line of reasoning? Why cannot I compute the dimension of $\Gamma_1$ with the formula for symmetric tensor products?
