Problem (Combinatorics):
A six-digit natural number is "sort-of-decreasing" if its first three digits are in strictly decreasing order and its last three digits are in strictly decreasing order. For example, 821950 and 631631 are sort-of-decreasing but 853791 and 911411 are not. Compute the number of sort-of-decreasing six-digit natural numbers.

Solution:
If three distinct digits are chosen from the set of digits $\{0,1,2, \ldots, 9\}$, then there is exactly one way to arrange them in decreasing order. There are $\left(\begin{array}{c}10 \\ 3\end{array}\right)=120$ ways to choose the first three digits and 120 ways to choose the last three digits. Thus the answer is $120 \cdot 120=\mathbf{1 4 4 0 0}$.

Final answer: 14400