# Number of functions from finite ordered set to finite ordered set

Lets say we have $a_1<a_2<a_3<a_4<a_5\in A$ and $b_1<b_2<b3\in B$

We alse have $f:A \rightarrow B$

How many such functions $f$, so that $f(a_1)\leq f(a_2)\leq f(a_3)\leq f(a_4)\leq f(a_5)$?

Being not very combinatorial I "bruteforced" solution, wrote python/sage script, which generated functions like (tuples (A,B):

((1, 1), (2, 1), (3, 1), (4, 1), (5, 1))

...

((1, 1), (2, 2), (3, 2), (4, 2), (5, 3))

...

((1, 3), (2, 3), (3, 3), (4, 3), (5, 3))

Which yields 21 possible solutions. However for $|A|=70$ and $|B|=30$ making crossproducts like $A\times B$, and then generating all permutations of length 70 is rather silly idea.

Any tips on how to aproach this?

This is equal to the number of non-negative integers solutions to $x_1+x_2+x_3=5$. Why? a function of this type is determined uniquely by the number of elements that are sent to $b_1,b_2$ and $b_3$.
So how many solutions does that equation have in non-negative integers. This is a standard stars and bars question, the answer is $\binom{5+2}{5}$.
In general if you want monotonic functions from $\{1,2,3\dots n\}$ to $\{1,2,3\dots k\}$ there are $\binom{n+k-1}{n}$ ways to do it.