How many independent components does a rank three totally symmetric tensor have in $n$ dimensions? 
How many independent components does a rank three totally symmetric tensor have in $n$ dimensions? 

Needed for the irrep decompositon of $3\otimes 3\otimes 3$ in here.
No idea where to start to prove this.
I did come up with a more clever way of figuring it out for $n=3$, but I don't think it can be generalized. In $3$ dimensions, a totally antisymmetric (rank three) tensor has one component. From here I just counted the components that are nonzero for a totally symmetric one. We have the $3$ diagonal components, obviously. Then there are the components of the form $iij$ (no sum). The $i$s run over $3$ values and $j$ over $2$. Since $2\cdot 3=6$ we have $1+3+6=10$ components. 
 A: By a 'rank three tensor' I think you mean an element of $\bigotimes^3V$. This is not what 'rank' means in mathematics. An element of $\bigotimes^kV$ is said to have order (or degree) $k$. 
A rank one tensor (or simple tensor) is a tensor of the form $v_1\otimes\dots\otimes v_m$. The rank of a tensor $T$ is the minimal number of rank one tensors needed to express $T$ as a sum.

Let $V$ be an $n$-dimensional vector space, then for any $k \in \mathbb{N}$, let $\operatorname{Sym}^kV$ denote the collection of symmetric order $k$ tensors on $V$; note that $\operatorname{Sym}^kV$ is a vector space. Let $v_1, \dots, v_n$ be a basis for $V$, then a basis for $\operatorname{Sym}^kV$ is given by 
$$\left\{\frac{1}{k!}\sum_{\sigma \in S_n}v_{\sigma(i_1)}\otimes\dots\otimes v_{\sigma(i_k)} \mid 1 \leq i_1 \leq \dots \leq i_k \leq n\right\}.$$
Therefore, $\dim\operatorname{Sym}^kV$ is equal to the number of non-decreasing sequences of $k$ integers in $\{1, \dots, n\}$. Let $x_1 = i_1 - 1$, $x_j = i_j - i_{j-1}$ for $j = 2, \dots, k$, and $x_{k+1} = n - i_k$. Note that the number of non-decreasing sequences of $k$ integers in $\{1, \dots, n\}$ is in one-to-one correspondence with the number of of solutions to $x_1 + \dots + x_{k+1} = n-1$ in non-negative integers. The latter number can be found using the stars and bars method from combinatorics; doing so, we see that
$$\dim\operatorname{Sym}^kV = \binom{(n-1)+(k+1)-1}{n-1} = \binom{n + k -1}{n-1} = \binom{n+k-1}{k}.$$
So every symmetric order $k$ tensor can be written uniquely as a linear combination of $\binom{n+k-1}{k}$ basis tensors. The coefficients in this linear combination are what you refer to as "independent components". In the case you worked out yourself, you had $n = 3$ and $k = 3$ in which case the number of independent components is 
$$\binom{3+3-1}{3} = \binom{5}{3} = 10.$$
A: In an d-dimensional space, the number n of components of a tensor of rank r (without specific symmetry) is
$$n=d^{r}$$
for totally symmetric tensor:
$$n=\binom{d+r-1}{r}=\frac{(d+r-1)!}{r!(d-1)!}$$
for totally anti-symmetric tensor:
$$n=\binom{d}{r}=\frac{d!}{r!(d-r)!}$$
