How many numbers of $4$ digits $x_1x_2x_3x_4$ satisfy $x_1\leq x_2 \leq x_3 \leq x_4$?

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.

from Hot Weekly Questions - Mathematics Stack Exchange