TITLE: How many numbers of $4$ digits $x_1x_2x_3x_4$ satisfy $x_1\leq x_2 \leq x_3 \leq x_4$?
QUESTION [6 upvotes]: How many numbers of $4$ digits $x_1x_2x_3x_4$ satisfy $x_1\leq x_2 \leq x_3 \leq x_4$?
What I've worked so far:
I have identified this situation with distributing $4$ equal objects among $9$ different boxes (digits from $1$ to $9$).
Therefore, solving the question proposed should be equivalent to solving the number of integer solutions of the following equation:
$$y_1+y_2+y_3+y_4+y_5+y_6+y_7+y_8+y_9=4\\
y_i\geq0 \quad \forall i\in\{1,2,...,9\}$$
The total solutions for this equations are: $$\text{CR}_{9}^{4}={4+9-1 \choose 4}={12 \choose 4}=\frac{12!}{4!\cdot8!}=11\cdot9\cdot5=495\text{ posibilities}$$
Is there any other approach you think it's more practical? Is this reasoning correct?
Clarifications:
If one distribution of the objets is $6292$ (box nº 6 is the first box where I distribute an object; box nº $2$ the second; box nº $9$ the third and box nº $2$ the fourth). The order in which boxes I distribute the objects doesn't really matter because the objects are identical, so I can rearrange the result to get $2269$ which is a combination that verifies the restriction given.

REPLY [3 votes]: Your solution is correct.  While I like your approach, here is another way of thinking about the problem.  
We can represent any four-digit number by placing four dividers in a row of nine ones, with the dividers placed to the immediate right of the digit that is included in the number.  For instance, we can represent $2269$ by 
$$1 1 \color{blue}{| |} 1 1 1 1 \color{blue}{|} 1 1 1 \color{blue}{|}$$
and $1234$ by 
$$1\color{blue}{|} 1 \color{blue}{|} 1 \color{blue}{|} 1 \color{blue}{|} 1 1 1 1 1$$
Each such number is completely determined by the positions of the dividers.  Since the first digit cannot be zero, the first of the $13$ symbols must be a $1$.  Thus, four of the remaining $12$ symbols must be dividers.  Choosing which four of those twelve positions will be filled with dividers determines the number.  Hence, as you found, there are 
$$\binom{12}{4}$$
four-digit numbers with non-decreasing digits.