Finding the dimension of the space of symmetric $r-$tensors Let $T^r(V^*)$ denote the space of $r-$tensors on $V$ a vector space of dimension $n$. Basis of $V$ is denoted $\{u_1, \dots u_n $} and basis of $T^r(V^*)$ is $\{\tilde{u}^{i_1} \otimes \dots \otimes \tilde{u}^{i_r}  : 1\leq i_1 \leq n, \dots , 1\leq i_r \leq n\}$ 
Let $\Sigma^r(V^*)$ be the (sub)space of symmetric $r-$tensors where if $\sigma \in S_r$ is a permutation in the symmetric group and $\alpha \in \Sigma^r(V^*)$, then: 
$$\alpha^\sigma (v_1, \dots , v_r) = \alpha (v_{\sigma(1)}, \dots , v_{\sigma(r)}) $$
What I am trying to do is determine the dimension of  $\Sigma^k(V^*)$. I know that there is a projection mapping $\mathcal{S}$ called the symmetrizer from $T^r(V^*)$ onto  $\Sigma^r(V^*)$ defined by: 
$$\mathcal{S}(\alpha) = \dfrac{1}{r!}\sum_{\sigma \in S_r} \alpha^\sigma$$. 
I also (from some searching) already know what the desired answer is: $$\text{dim}\Sigma^k(V^*) = \binom{n + r -1}{r} = \dfrac{(n + r -1)!}{r! (n-1)!}$$
However, I am unable to derive the general expression myself. My current approach, which has worked for $V = \mathbb{R}^4$ and $r =2, 3$ has been to consider the basis of $T^r(V^*)$ and see which elements give the same element of $\Sigma^k(V^*)$ upon symmetrization. For example for the $r=2$ case, I notice that $\mathcal{S}(\tilde{e}^1 \otimes \tilde{e}^2) = \mathcal{S}(\tilde{e}^2 \otimes \tilde{e}^1)$ ; $\mathcal{S}(\tilde{e}^1 \otimes \tilde{e}^3) = \mathcal{S}(\tilde{e}^3 \otimes \tilde{e}^1)$ ; $\mathcal{S}(\tilde{e}^1 \otimes \tilde{e}^4) = \mathcal{S}(\tilde{e}^4 \otimes \tilde{e}^1)$ ; $\mathcal{S}(\tilde{e}^3 \otimes \tilde{e}^2) = \mathcal{S}(\tilde{e}^2 \otimes \tilde{e}^3)$; $\mathcal{S}(\tilde{e}^4 \otimes \tilde{e}^2) = \mathcal{S}(\tilde{e}^2 \otimes \tilde{e}^4)$ ;  and $\mathcal{S}(\tilde{e}^3 \otimes \tilde{e}^4) = \mathcal{S}(\tilde{e}^4 \otimes \tilde{e}^3)$. To these $6$ terms, I add the symmertrizations of $\tilde{e}^i \otimes \tilde{e}^i$ for $i \in \{1,2,3,4\}$ to get dimension 10, as required. 
 A: I think I have figured out the answer, so I'm going to post a solution for it to be critiqued and hopefully for the benefit of someone who runs into the same problem in the future. I realized I was having problems because I wasn't quite familiar with the classical combinatorics result this mirrors. 
My test cases made it clear that the solution would be bounded below by $\dfrac{n!}{n-r)!}$, that is, the number of distinct $r-$sequences without repetitions that can be extracted from $\{1,2, \dots, n\}$. What remains to be determined are how many additional terms there are when we permit repetitions. If consider $r-$sequences $I = \{1,2,2,1,\dots, 2\}$ and $J = \{2,1,1, \dots, 2,1\}$ where $I$ and $J$ are both composed of purely $1s$ and $2s$ with the same count of each only with the order rearranged i.e. $I$ and $J$ are permutations of each other. 
In general, we find that $\mathcal{S}(\tilde{e}^I) = \mathcal{S}(\tilde{e}^J)$ if $I$ and $J$ are $r$-sequences that permutations of each other.
So, our problem has been reduced to: How many distinct ways of selecting/choosing $r$ elements from $n$ elements, without caring for order or repetitions are there? The answer is: $\binom{n +r -1}{r}$
This is quite a foundational result that can be found in any introductory discrete mathematics book. An example would be Section 9.6 of Discrete Mathematics with Applications (4th Ed.) by Susanna S. Epp. 
