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# A “descending number” is a three-digit number, such that the units digit is less than the tens digits and the tens digit is less than the hundreds digit. what is the probability that a three-digit number chosen at random is a “descending number”?

Fix the first digit to be 2. How many two-digit descending numbers can you make? Only one, and that would be $\mathbf{210}$. Now fix the first digit to be 3. How many two-digit descending numbers can you make? There are three, and these are $310, \mathbf{320}, \mathbf{321}$. Next, if the first digit is 4, then there are six possible descending numbers, $410, 420, 421, \mathbf{430}, \mathbf{431}, \mathbf{432}$. You might start seeing a pattern here. If the first digit is $n$, then each choice of $n$ from $\{1,2,\ldots,9\}$ contributes $n-1$ more possible two-digit permutations that are descending. As the pattern continues, you'll find that the total number of descending three-digit numbers is $\displaystyle\sum_{n=1}^9\frac{n(n-1)}2=120$ Meanwhile, there are $900$ possible three-digit numbers that can be randomly chosen (100 through 999), so the probability you're looking for is $\dfrac{120}{900}=\dfrac2{15}$.