TITLE: The number of non-decreasing sequences of digits
QUESTION [5 upvotes]: What's the number of n-digit natural numbers, in which digits are in non-decreasing order?
The answer is $ n+8 \choose 8$, but I don't understand how to get this score - could anyone try to explain it to me?

REPLY [5 votes]: The digit $0$ can appear in the number only if $n$ is equal to $1$. So, if $n=1$, then the answer is $10$ different numbers. If $n>1$, then the numbers can contain only digits from set $A = \{1,2, ... ,8,9\}$. $0$ cant't appear in the number when $n>1$ becouse you couldn't then construct n-digit natural number, in which digits are in non-decreasing order.
Let's focus on the situation where $n > 1$. So, we have non-decreasing n-digit number $a = d_1d_2 ... d_{n - 1}d_{n}$, where the digits are elements of the set $A = \{1,2, ... ,8,9\}$. The numer of n-digit natural numbers, in which digits are in non-decreasing order is a $n$-combination with repeated elements chosen within the set $A$. Why? Because if you just take any $n$-combination with repeated elements from the set $A$ (for example, $\{2,4,4,2,5\}$) and sort it ascending then you will get one of the desired number ($\{2,2,4,4,5\}$). Each combination represents exactly one searched number. Thus, Simply find the total number of $n$-combination with repeated elements from the set $A$.
The number $C^{{\prime}}_{{a,n}}$ of the $n$-combinations with repeated elements within $a$-element set is given by the formula:
$$
C^{{\prime}}_{{a,n}}=\binom{a+n-1}{n}.
$$
Thus, for $n>1$ the solution is equal to the number of:
$$
C^{{\prime}}_{{9,n}}=\binom{9+n-1}{n}=\binom{n+8}{n}.
$$
Widely known is the following rule :
$$
\binom{m}{k}=\binom{m}{m-k}.
$$
Using the given above rule let's give final answer for $n>1$:
$$
\binom{n+8}{n}=\binom{n+8}{n+8 -n}=\binom{n+8}{8}
$$