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