Question

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”?

Answer 1

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
`(120)/(900)=\frac2{15}`.