Count the number of functions Let $m$ and $n$ be two integers such that $m \geq n \geq 1$. Count the number of functions $f : \{1,2,3=...,n \} \rightarrow \{1,2,3=...,m \}$ such a that $f$ is strictly increasing. 
My attempt:
case 1: $m=n=1$,  only 1 , $f(1)=1$ 
case 2: $m>n=1$, $m$-many, as $f(1)$ can take any of the $m$-values in the range
case 3 : $m=n>1$, identity function, only $1$
case 4: $m>n>1$, $1$ can be mapped to $k_1 , 1 \leq k_1 \leq (m-n+1) $... i don't know how should I proceed and put the rest of the calculation into a formula. Any hints will be appreciated...

If we choose $n$ elements in $\{1, 2, \cdots, m\}$, we can correspond the case to each $f$. 
For example, if $m=3, n=5$, in $\{1, 2, 3, 4, 5\}$ and choose 3 elements in that set, for example, 2, 3, 5, then the corresponding $f$ is defined as $f(1)=2, f(2)=3, f(3)=5$. 
Therefore the number of functions satisfying the strictly increasing condition is $\binom{m}{n}$ if $m\geq n$ and $0$ otherwise. 
