a). Number of Strictly increasing functions $f:\{1,2,3,\ldots,n\}\to\{1,2,3,\ldots,m\}$ where $m\geq n \geq 1$

A function $f:\{1,2,3,\ldots,n\}\to \{1,2,3,\ldots,m\}$ is determined by how the values $f(1),f(2),f(3),\ldots,f(n)$ are assigned. Since $f$ is a strictly increasing function and for a strictly increasing function, each value in the domain is mapped to a distinct element in the co-domain.

$f(1)<f(2)<f(3)<\ldots<f(n)$

So the number of ways to select $n$ elements from a set with $m$ elements is - $\Large\binom{m}{n}$

Once we have selected these elements, there is only one way to assign them -assigning the smallest element in the range to be f(1), the next smallest to be f(2), and so on. Hence, the number of strictly increasing functions is- - $\Large\binom{m}{n}$

b). Number of Non Decreasing functions $f:\{1,2,3,\ldots,n\}\to \{1,2,3,\ldots,m\}$ where $m \geq n \geq 1$

Since $f$ is a non-decreasing function, the function is completely determined by how many values of $j\in \{1,2,3,\ldots,n\}$ are assigned to equal $k\in \{1,2,3,\ldots,m\}.$

For example - consider non-decreasing functions

$f:\{1,2,3,4,5\}\to\{1,2,3,4,5,6,7\}$

Then we have - $x_1+x_2+x_3+x_4+x_5+x_6+x_7 = 5$

And suppose it is defined as - $f(1)=f(2)=3,f(3)=4,f(4)=f(5)=7$

so $x_1 = 0,x_2 = 0,x_3 = 2,x_4 = 1,x_5 = 0,x_6 = 0x_7 = 2$

Let $x_k, 1\leq k \leq m$, be the number of values of $j\in \{1,2,3,\ldots,n\}$ such that $f(j)=k$ . Then

$x_1+x_2+x_1+\ldots +x_m = n$

which is an equation in the nonnegative integers.

It's solution is given by - $\Large\binom{m+n-1}{n}$

So, number of non decreasing functions $f:\{1,2,3,\ldots,n\}→\{1,2,3,\ldots,m\}$ where $m\geq n \geq 1 = \Large\binom{m+n-1}{n}$