Number of cover relations on partition $L(m,n)$ is the set of partitions with length m and each part is in the set ${0,...n}$. Two partitions $a$ and $b$ in $L(m,n)$  with $a=(a_1,...,a_m)$ and $b=(b_1,...,b_m)$ are in relation:
$a\leq b$ if $a_i\leq b_i$ for all $i$ and $a_i<b_i$ for at least one $i$.
$c(n,m)$ is the number of pairs $(a,b),  \; a,b \in L(m,n)$ with $a\leq b$ and no partition $c$ with $a\leq c \leq b$.
I need to show that:
$$c(m,n)=\frac {(m+n-1)!}{(n-1)!(m-1)!}$$
I already have the hint to watch the $a_1$ but do not progress to the solution. Any help apreciated! Thanks.

That's very odd notation for a non-reflexive relation, but never mind.
What you're trying to count are pairs $(a, b)$ which agree on every part except one, and for that one part the difference is $1$. So either $a_1=b_1=k \le n$, in which case the remaining rows have $c(m-1,k)$ possibilities; or $a_1= b_1-1 = k<n$, in which case the remaining rows are equal to each other and have $|L(m-1,k)|$ possibilities.
