Let $n\ge 2$ be an integer. The symmetric group $S_3$ acts on the set $M_n$ of polynomials in $\mathbb{C}[x_1,x_2,x_3]$ whose monomials are of the form $x_1^{a_1}x_2^{a_2}x_3^{a_3}$ with $0\le a_i\le n$ in the obvious way.
Is there a simple way to describe how $M_n$ is decomposed as a sum of the three irreducible representations of $S_3$ ?
For example,
$$M_2=\langle 1\rangle\oplus \langle x_1x_2x_3\rangle\oplus \langle x_1+x_2+x_2\rangle\oplus \langle x_1-x_2,x_1-x_3\rangle\oplus\langle x_1^2+x_2^2+x_3^2\rangle\oplus \langle x_1^2-x_2^2,x_1^2-x_3^2\rangle\oplus \langle x_1x_2+x_2x_3+x_1x_3\rangle\oplus \langle x_1x_2-x_2x_3,x_1x_2-x_1x_3\rangle,$$ but this decomposition cannot be directly generalized even for $M_3$.
Let us compute the character of the representation of $M_n$. We need to compute the trace of the action of the identity element, of $(12)$ and of $(123)$. These permute the monomials in $M_n$, so we are just counting how many monomials each of these permutations fix.
First: the identity element of course fixes all monomials, so its trace is $\binom{n+3}{n}$, which is the dimension of $M_n$.
Second: the element $(12)$ fixes the monomials $x_1^ax_2^bx_3^c$ with $a=b$ and $a+b+c\leq n$, and there are as many of these are there are even non-negatve numbers not larger than $n$, that is $\lfloor n/2\rfloor$.
Third: the element $(123)$ fixes the monomials $x_1^ax_2^bx_3^c$ with $a=b=c$ and $a+b+c\leq n$, of which there are $\lfloor n/3\rfloor$.
If we write $n=6q+r$ with $0\leq r<6$, then the trace of $(12)$ is $3q+\lfloor r/2\rfloor$ and that of $(123)$ is $2q+\lfloor r/3\rfloor$. If you know the character table of $S_3$, then now decomposing $M_n$ is a matter of solving a linear system.