Case 1: Indicates that function is non-decreasing. Which means that, from the co-domain of the first $m$ numbers, we have to pick elements such that all $n$ numbers get a value assigned to them. This assignment is a bijection. To understand this let's assume $k$ numbers are the ones assigned. These k numbers have a single ordering to be assigned to the $n$ numbers in the domain.
An example for $n=4$ and $m=5$, if you know that in the image of $f$ there are two $1$s, two $4$s . Then the only function with that property (which is non-decreasing) is:
\begin{align}
f(1)=1\\
f(2)=1\\
f(3)=4\\
f(4)=4\\
\end{align}
Thus, we can see that this is same as finding the number of solutions for : $\,x_1+x_2+\cdots+x_m=n$. Which is: $$ \bbox[yellow,5px,border:2px solid red]
{\;\binom{n+m-1}{n}
}
$$
Case 2: Indicates that the function is strictly increasing. Thus the numbers in the range are clearly distinct. This is simply choosing n numbers from the set of m numbers. Which is:
$$ \bbox[yellow,5px,border:2px solid red]
{\;\binom{m}{n}
}
$$
Case 3: Indicates that function is non-increasing. This is similar to case 1 and the same bijection in the reverse order applies. So the answer would be same as Case 1 : $\color{blue}{\binom{n+m-1}{n}}$
Case 4: Symmetric case of Case 2 : $\color{red}{\binom{m}{n}}$
Hope it helps..
