Case $Sym^n(K^m)\cong K^6$ for $(m,n)$ 
Let $K$ be a field. Find all the pairs $(m,n)\in\mathbb{N}_{>0}\times\mathbb{N}_{>0}$ such that $Sym^n(K^m)\cong K^6$

So I just need to find all the $(m,n)$ such that $\dim(Sym^n(K^m))=6$.
We've seen in lcture that $\dim(Sym^n(K^m))=\binom{m+n-1}{n}$
So we have: $\binom{m+n-1}{n}=6 (=3!)$
Since we have two variables in one equations, I just made few attempts, finding $(m,n)=(3,2)=(2,5)$
Do you think I should do something else, or that I could find a general formula to find the values?
 A: You are looking for binomial coefficients that are equal to$~6$. Obviously those coefficients are never the extreme ones $\binom n0$ or $\binom nn$, since all of those are equal to$~1$.
Binomial coefficient grow as one goes down Pascal's triangle. Indeed Pascal discovered that each coefficient not on the border is the sum of two coefficients in the previous row*. So if the coefficient is to be (at most)$~6$, those other two need to be less than$~6$. It is now a fairly simple matter to classify all binomial coefficients $1<\binom ln\leq 6$, and one finds that $(l,n)\in\{(4,2),(6,1),(6,5)\}$ describes the complete solution. Then you can put $m=l-n+1$ in each solution.
So it seems that you are overlooking (just) the most obvious case $(m,n)=(6,1)$.
*This is not quite accurate. Indeed, Pascal defined the entries of his triangle like that, and later showed them to be equal to binomial coefficients (more precisely he says "d'autres en ont déjà traité, comme Hérigone, outre que la chose est évidente d'elle même").
