# The Dimension of the Symmetric $k$-tensors

I want to compute the dimension of the symmetric $k$-tensors. I know that a covariant $k$-tensor $T$ is called symmetric if it is unchanged under permutation of arguments. Also, I know that the dimension of covariant $k$-tensors is $n^k$ but how can I eliminate non-symmetric the cases? I found this but I could not get the intution. Also, this blog post answers my question but I don't see why we put | between different indices. Any concrete example would also help such as the symmetric covariant 2-tensors in $\mathbb{R^3}$, as I asked in this thread.

-
@Thomas They are equal indeed. –  Erno Nemecsek Dec 25 '11 at 13:57
While tensor (fields) are used extensively in differential geometry, your question is not specific for differential geometry. You might want to consider (re)tagging it with 'tensors', 'tensor-products', 'multilinear-algebra'. –  user20266 Dec 25 '11 at 13:59
a basis for symmetric tensors, say $\otimes_1^r V$ with $\{v_1,...,v_n\}$ a basis for $V$, is given by $\{v_{i_1}\otimes ... \otimes v_{i_r} \ | \ 1\leq i_1\leq...\leq i_r\leq n\}$. so you must count the number of non-decreasing sequences (repetitions allowed) of length $r$ with entries in $[1,n]$. I've always heard the method of counting these referred to as stars and bars, ie counting the number of multisets of size $r$ with entries from $[1,n]$, and the answer you get is ${n+r-1\choose r}$.
Hint.... take $\{i_1\le \cdots \le i_k\}$ then shift this set in the following way $\{i_1+0<i_2+1<i_2+2< \cdots < i_k+k-1\}$ Now how big is $\{i_1,...,i_k+1-1:1\le i \le n\}$? –  Squirtle Apr 24 '13 at 19:15