how many monotonically increasing functions are in $\Omega = \{ \space f \mid f: A \rightarrow B\space \}$ Given $A=\{1,2,3..,n\}$,  $B=\{1,2,3,..,m\},$$\space \space $while $5<m,n$ $\space$ and $\space \Omega = \{ \space f \mid f: A \rightarrow B\space \}$
How many functions are monotonically increasing in $\Omega$ while $m>n$?
My answer is $m-n+1$. 
Basic example is $A=\{1,2,3,4,5\}$ and $B=\{1,2,3,4,5,6\}$.
hence:
$$f_1(1) = 1, \space \space f_1(2) = 2, \space \space f_1(3) = 3, \space \space f_1(4) = 4, \space \space f_1(5) = 5$$
$$f_2(1) = 2, \space \space f_2(2) = 3, \space \space f_2(3) = 4, \space \space f_2(4) = 5, \space \space f_2(5) = 6$$
So we get 2 functions, and while the diffrence between $m$ and $n$ getting bigger, the result's getting bigger.
Am I missing somthing?
 A: $${m}\choose{n}$$
Or it is probably easier to think about the answer as ${m}\choose{m-n}$, but that is equal to ${m}\choose{n}$ of course. We can construct all of the functions of $\Omega$ in the following way. Think of $m$ "blanks". So if $m=5$ for example, we have 5 blanks, which I will denote by underscores ( _ ):
_ _ _ _ _
Then if $n=2$ for example, mark out $m-n = 3$ of these blanks with x's. Here's one way to do this:
x _ x x _
Note that this is merely one way to put x's in. There are ${m}\choose{m-n}$ total ways to do it.
Then fill in the remaining blanks with $1,2,\ldots, n$ (in order) until there are no blanks left:
x 1 x x 2
This is the function $f(1) = 2,\ f(2) = 5.$ Every such function can be uniquely constructed in this way. Therefore there are ${m}\choose{m-n}$ such functions.
A: In question, number of monotonically increasing functions from $A$ to $B$ is asked, NOT that of "strictly increasing".
The question can be thought as number of ways of slicing $A$ to $m$ pieces while keeping order of its elements. So we have to make $m-1$ cuts to cut $A$ to $m$ pieces such that if $a$ is in $kth$ slice, $f(a)= k$. Imagine slices as follows:
_ /_ /_ /_ ... / _ 

Here there are $m$ slots (_) and $m-1$ slicers (/). We are after the number of ways to order this slicers and the elements of A. For example suppose $m = 4$ and  $A = \{1,2,3\}$ One possible slicing would be:
1,2/ _ /_ / 3

which means: $f(1)=f(2)=1, f(3)=4$
So we have $n$ elements from $A$ and $m-1$ slicers to order. To place all these, now imagine we have $n+m-1$ slots. If we choose $m-1$ slots among these to place the slicers, there will be one way to place elements from $A$, since we must keep the order among them.
So the number of ways to place all n+m-1 slicers and elements of A is  ${n+m-1\choose n-1}$
