Smart counting

Let $a_{k}$ for $6 \le k \le 10$ be the number of (unspecified) elements of $A$ that $f$ maps to element $k$ of $B.$ Then there is a one-to-one correspondence between the set of non-decreasing functions $f$ and the set of non-negative integer solutions to the equation

$a_{6} + a_{7} + a_{8} + a_{9} + a_{10} = 5,$ (A).

(This is because for any solution to (A), there will be only one way to then form the mapping $f,$ and any non-decreasing mapping corresponds to a unique solution of (A). For example, the solution $a_{6} = 0, a_{7} = 2, a_{8} = a_{9} = a_{10} = 1$ corresponds to the mapping $f(1) = f(2) = 7, f(3) = 8, f(4) = 9$ and $f(5) = 10.$ )

This is now a stars and bars calculation, and according to Theorem two in the link the number of solutions to (A), and hence the number of non-decreasing functions $f$ , is

$\dbinom{5 + 5 - 1}{5} = \dbinom{9}{5} = \boxed{126}.$

Let $N(n,m)$ denote the number of such non-decreasing functions $f:\{1,2,\ldots, n\} \to B$ such that $f(n)=m$ ( $m \geq 6$ ). Then because of non-decreasing property of $f$ , we have $f(n)\geq f(n-1)$ . Hence, $N(n,m)=\sum_{k=6}^{m} N(n-1,k)$ with the initial conditions $N(1,m)=1, N(n,6)=1\forall m \geq 6,n$ . The above recursion can be solved either analytically or computationally ( we just need to fill up a $5 \times 5$ table). We require $\sum_{m=6}^{10}N(5,m)=126$