How do I count the numer of degrees of freedom in a totally symmetric tensor with $m$ lower (or upper) indices taking values in $\{ 1,2,3 \}$?
I know that the result is $$dim(m,0)=\frac{(m+2)(m+1)}{2}$$
For a symmetric tensor that takes values $\{ 1,2 \}$ I get the counting: indices $1$ can be $m,m-1,...,1,0$ and therefore $dim=m+1$.
For three possible values my assignment says we can split m objects into three sets (first splitting m objects into two sets, and then splitting the second part in two again). Can someone, in more detail, explain the counting or splitting in sets?
Values in the tensor are indexed by $m$-tuples $(a_1,a_2...a_m)$, where $a_p \in \{1..k\}$ (in your case k = 3). The symmetry imposes an equivalence relation on these tuples, where two tuples are equivalent if they contain the same values, regardless or order. For example, in a totally symmetric $4$-tensor, the index $(5,1,2,2)$ gives the same value as the index $(1,2,5,2)$. Now from each such equivalence class of tuples we select a "canonical" member which will represent the class.
This will be the $m$-tuple for which all the values are ordered. For our example this gives $(1,2,2,5)$. We need to count the $m$-tuples $(a_1,a_2...a_m)$ such that $(j<s)\rightarrow (a_j \leq a_s)$ . For $a_p \in \{1..k\}$ this gives us exactly $C^{k+m-1}_{m}$:
We make a bijection taking $(a_1,a_2,a_3,...a_m)$ to $(a_1,a_2+1,a_3+2,...a_m+(m-1))$. If $(a_1,a_2,a_3,...a_m)$ is (non-strictly) ordered then $(b_1,b_2,b_3,...b_m) =(a_1,a_2+1,a_3+2,...a_m+(m-1))$ is strictly ordered an each of its terms is distinct.
As we have $a_p \in \{1..k\}$, we have $b_p \in \{1,..k+m-1\}$. The counting of $m$ distinct ordered terms with values from $1$ to $k+m-1$ is done by combinations as $C^{k+m-1}_{m}$. By the above bijection, this is equivalent to counting $m$ (not-necessarily distinct) ordered terms with values form $1$ to $k$, which index the degrees of freedom of a totally symmetric tensor with $m$ indices in dimension $k$.
Setting $k=3$ we get $C^{3+m-1}_{m} = C^{2+m}_{m} = \frac{(m+2)!}{m!2!} = \frac{(m+1)(m+2)}{2}$.