Decomposition of a representation of $S_3$ on monomials 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 $M_3$. 
 A: Let $V=\langle x_1,x_2,x_3 \rangle$  be  the natural representation of $S_3.$
Then 
$$
M_n={\rm Sym}^0(V) \oplus {\rm Sym}(V) \oplus \cdots  \oplus {\rm Sym}^{3n}(V),
$$
where ${\rm Sym}^k(V)$ is the symmetric power of the vector space $V.$ All that you need is to decompose ${\rm Sym}^k(V)$ into  irreducible representations. There are $3$ such representations - one dimensional  trivial $V_0$ and  alternating $U$ representations and  two dimensional standard  representation $W.$  Note that $V=V_0 \oplus W.$  We have something like as (up to coefficients)
$$
{\rm Sym}^k(V)={\rm Sym}^k(V_0 \oplus W)\cong  V_0 \oplus  {\rm Sym}(W) \oplus {\rm Sym}^2(W) \oplus\cdots \oplus  {\rm Sym}^k(W).
$$
Therefore  you need to decompose the symmetric  power of standard representation $W$ of $S_3.$ But it is the Exercise 1.12 of  the Fulton, Harris' book. I hope they plase  an answer at the end of the book. 
After that you should put together everythings.
A: 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.
