I am trying to solve a problem in algebric combinatorics, where i want to prove that for integers $m,n$ and pairs ($\lambda$ , $\mu$) of integer partitions with a maximum of $m$ non zero parts which are less or equal to $n$ and where $ \lambda \lessdot \mu$ (which denotes that $\lambda \neq \mu$ and there exist no $x$ with $\lambda < x < \mu$) the following function counts those pairs: $$c(n,m) = \frac{(m + n - 1)!}{(m - 1)!(n - 1)!}$$ I know that this problem is related to counting lattice paths on a grid with side lengths $m$ and $n$ as there exists a bijection from the integer partitions in question onto the paths on that lattice. Further we can count those paths with ${m + n}\choose{n}$ and display each of these paths as a corresponding young diagram whereas $\lambda \lessdot \mu$ equals to the young diagram of $\lambda$ being covered by the young diagram of $\mu$.
Edit: What I tried now is to differentiate between different values for $\mu_1$ being the highest integer in the partion, so that $$ \binom{m + n}{n} = \sum_{i = 0}^{n}\binom{(m - 1) + i}{m - 1}$$ where each summand equals to finding the count of lattice paths in a grid with side lengths $i$ and $(m-1)$ where $i$ is the fixed number $\mu_1$ in $\mu = (\mu_1,...,\mu_n)$. Further $$c(n,m) = \binom{m + n - 1}{m - 1} n = \frac{(m + n - 1)!}{(m - 1)!(n)!}n = \frac{(m + n - 1)!}{(m - 1)!(n - 1)!}$$
this is $n$ times the count of paths on a grid with sides of length $m - 1$ and $n$. Am i able to follow from fixing $\mu_1$ to counting on that smaller grid somehow? Any suggestions how to continue from here?