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