I want to know how to calculate number of non decreasing functions from one set to another set. Let $A=\{1,2,3,\ldots,10\}$ and $B=\{1,2,3,\ldots,25\}$
Please tell me an easy method to calculate the number of non decreasing functions from set $A$ to $B$.
I can understand permutations and combinations, but I need an explanation which sticks to a basic level.
Let $[n] = \{1, 2, 3, \ldots, n\}$. Then $A = [10]$ and $B = [25]$.
A non-decreasing function $f: A \to B$ is completely determined by the number of times each element of $B$ appears in the range. For instance, if $1, 3, 5, 7, 9, 11, 13, 17, 19$ each appear once in the range, then
\begin{align*}
f(1) & = 1 & f(6) & = 11\\
f(2) & = 3 & f(7) & = 13\\
f(3) & = 5 & f(8) & = 15\\
f(4) & = 7 & f(9) & = 17\\
f(5) & = 9 & f(10) & = 19
\end{align*}
If, instead, $3$ appears in the range four times, $14$ appears in the range three times, $19$ appears in the range twice, and $25$ appears once, then
\begin{align*}
f(1) = f(2) = f(3) = f(4) & = 3\\
f(5) = f(6) = f(7) & = 14\\
f(8) = f(9) & = 19\\
f(10) & = 25
\end{align*}
Let $x_k$ denote the number of occurrences of $k$ in the range of $f$. Then the number of non-decreasing functions $f: A \to B$ is the number of solutions of the equation
$$x_1 + x_2 + x_3 + \cdots + x_{25} = 10 \tag{1}$$
in the non-negative integers. A particular solution of equation 1 corresponds to the placement of twenty-four addition signs in a row of ten ones. For instance,
$$+ + 1 + + 1 1 + + + 1 + + 1 + 1 + + + + 1 + + + + 1 + + + + + + 1$$
corresponds to the solution $x_1 = x_2 = 0$, $x_3 = 1$, $x_4 = 0$, $x_5 = 2$, $x_6 = x_7 = 0$, $x_8 = 1$, $x_9 = 0$, $x_{10} = x_{11} = 1$, $x_{12} = x_{13} = x_{14} = 0$, $x_{15} = 1$, $x_{16} = x_{17} = x_{18} = 0$, $x_{19} = 1$, $x_{20} = x_{21} = x_{22} = x_{23} = x_{24} = 0$, $x_{25} = 1$. Thus, the number of solutions of equation 1 in the non-negative integers is
$$\binom{10 + 24}{24} = \binom{34}{24}$$
since we must select which $24$ of the $34$ symbols (ten ones and $24$ addition signs) will be addition signs.
Well, if $S$ is an ordered finite set, we can suppose, that there is $n$ such that $S$ is isomorphic (in the mean of ordering) to the set $\{1,2,3,\ldots\}$. So, we can suppose $A=\{1,2,3,\ldots,n\}$ and $B=\{1,2,3,\ldots,m\}$. Consider first the case, when $n$ is not greater than $m$ If we would have to count the strictly increasing functions from $A$ into $B$, we would have just to count the subsets of $B$ containing $n$ elements, because a strictly monotone increasing map from $A$ into $B$ is determined by its image.
To step further, we have to notice, that if $b_1,\ldots,b_n$ is a non-decreasing sequence in $B$, then $b_1, b_2+1, \ldots, b_n+n-1$ is a strictly increasing sequence in $\{1, 2, \ldots, m+n-1\}$. We can check, if two non-decreasing sequence isn't the same, the derived increasing sequence isn't the same too. On the other hand, if $c_1,c_2,\ldots,c_n$ is a strictly increasing sequence in $\{1, 2, \ldots, m+n-1\}$ then $c_1,c_2-1,\ldots,c_n-n+1$ is a non-decreasing sequence in $B$. Different increasing sequences in $\{1, 2, \ldots, m+n-1\}$ determine different non-decreasing sequences in $B$. With this we outlined, how to prove, that number of the non-decreasing functions from $A$ in $B$ is the same as the number of increasing sequences in $\{1,2,\ldots,m+n-1\}$ having $n$ members.
The other case, when $n$ is greater than $m$ is a bit more difficult.
