Note: I am aware a solution already exists to this question on SE, but I am questioning why the approach I am using here is incorrect / incomplete.
The only contents of a container are 10 disks that are each numbered with a different positive integer from 1 through 10, inclusive. If 4 disks are to be selected one after the other, with each disk selected at random and without replacement, what is the probability that the range of the numbers on the disks selected is 7?
I am aware of this solution to solve the problem using combinatorics: The total possible outcomes are $\binom {10}{4}$ and we are looking for the following:
Case 1: 1, 8, 2 from {2-7}
Case 2: 2, 9, 2 from {3-8}
Case 3: 3, 10, 2 from {4-9}
Which is basically $3 * \binom{6}{2} = 45.$
However, I wanted to solve this using the 'probability approach'. So for example, for the first case there are 30 possible sets of numbers (e.g. 1823, 1824, etc.). Then there are 24 possible ways these numbers can be arranged. Then there are 3 cases. So the probability according to my understanding should be:
$\frac{1}{10} * \frac{1}{9} * \frac{6}{8} * \frac{5}{7} * 24 * 3 = 3/7$
However, the answer provided is $3/14$. I seem to be double-counting something, but can't figure out what. Help would be appreciated.