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 \(\dfrac{120}{900}=\dfrac2{15}\).
