Best proof for the number of independent components of a totally (anti) symmetric tensor I'm asking this question for future reference. Given a totally antisymmetric and a totally symmetric tensor having $k$ indices living in $n$ dimensions. Denote the antisymmetric and symmetric tensor by $A_{i_1\dots i_k}$ and $S_{i_1\dots i_k}$ respectively. Without these symmetries would have $n^k$ independent components. Because of their (anti) symmetricity each of the components can be written such that the indices appear in ascending order. For example $A_{321}=-A_{123}$. The independent components can be written as a list where the indices of each element are ascending. The antisymmetric tensor has the additional constraint that no two indices can be the same.
\begin{align}
\text{indep}(A)&=\left\{A_{i_1\dots i_k}\middle|1\leq i_1<\dots<i_k\leq n\right\}\\
\text{indep}(S)&=\left\{S_{i_1\dots i_k}\middle|1\leq i_1\leq\dots\leq i_k\leq n\right\}
\end{align}
Now it is known that the number of independent components are given by
\begin{align}
\#(A)\equiv\big|\,\text{indep}(A)\,\big|&=\pmatrix{n\\k}\\
\#(S)\equiv\big|\,\text{indep}(S)\,\big|&=\pmatrix{n+k-1\\k}
\end{align}
My question is what are nice ways to proof this last statement (both for the symmetric/antisymmetric case). With  nice I mean either easy to remember, intuitive or elegant. The antisymmetric case has been proven here and both cases have also been proven in this paper. The reason I'm asking this question is to have these facts nicely together and also to get some more proofs to make it easier to remember. Also I found a nice proof myself.
 A: The antisymmetric case can be rephrased as 'I have $n$ digits. In how many ways can I group them in groups of $k$'. This has the answer $\pmatrix {n\\k}$.

The symmetric case is also in the form of an binomial coefficient so this invites us to the idea of writing the components of a symmetric $n,k$ tensor as an antisymmetric $n+k-1,k$ tensor. Let's call it $A'$. The difference between the antisymmetric/symmetric components is that the symmetric components can have repeating indices while the antisymmetric components can't. Let's construct the following map from $\text{indep}(S)\rightarrow\text{indep}(A')$: if $S_{i_1\dots i_k}$ has no repeating indices then map it directly to $A'_{i_1\dots i_k}$. If $S$ does have repeating indices we record the places of repeating indices. Then delete all repeating indices, leaving us with a couple indices to be filled. Fill these empty indices with the locations but add  $n$ to each of them. 

For example let's take a look at $S_{1,1,1,5,9}$ where $k=5,n=9$. In $S_{1,1,1,5,9}$ we have repeating indices at location 1 and 2. If we delete all repeating indices we get $A'_{1,5,9,\cdot,\cdot}$ with two spots left. We fill these spots with $9+1=10$ and $9+2=11$ to 'remember' where we deleted indices. This gives $A'_{1,5,9,10,11}$. Another example: $S_{5,5,7,7,7}\rightarrow A'_{5,7,10,12,13}$. This new tensor has dimension $n+k-1$ since $k-1$ is the largest position at which a double index can occur $n+k-1$ is the largest index that will occur. It also stricly increasing indices so it can be mapped to an antisymmetric tensor and correspondingly it has $\pmatrix{n+k-1\\k}$ components. For any antiysymmetric tensor we can reverse this map so it is a bijection. So we can conclude that our original symmetric tensor has the same number of components as this new tensor. So $\#(S)=\pmatrix{n+k-1\\k}$
