Confusion in one of the combinatorics/probability problem, with my approach for its solution. 
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?

My approach: (Using the concept of conditional probability and combinatorics)
Possible ranges: 1-8, 2-9, 7-10 (Total 3)
Now, taking one range at a time, (say 1-8):
Hence, among the 4 disks to be selected, the probability of selecting the first disk will 1/10 (for 1).
Similarly, 1 is selected, hence second place could be filled by 8. Prob(second place is 1/9).
Now, since out of range 1-8 1 and 8 are selected, therefore we have only 6 choices (out of the remaining 8) for third place. Henceforth, only 5 choices (out of 7 choices) for fourth place.
Combining the probability of all 4 places, (1/10)(1/9)(6/8)(5/7).
Now, the number in four places can be arranged in 4! ways. Also, apart from the (1-8) range, there is also other two range of numbers.
Therefore, combining all, the required probability should be (3) x (1/10)(1/9)(6/8)(5/7) x (4!).
But my answer is wrong. My answer is double of actual answer.
 A: I'm not exactly sure where you're double counting.  The method you've elected to use is rather confusing.
You've got all the pieces, but there's a simpler way to put them together.  There are, as you said, $3$ possible ranges.  There are $\binom62$ ways to choose $2$ numbers in the range other than the endpoints, so there are $3\binom62$ possible choices.  There are $\binom{10}4$ equiprobable ways to choose $4$ numbers, so the probability is $$\frac{3\binom62}{\binom{10}4}=\frac3{14}$$
EDIT
I'm having difficulty pinpointing your error.  As I said, the method is confusing.  I think the problem arises when you compute the probability of choosing $2$ numbers from the middle of the range.  You say the probability is $\frac{6\cdot5}{8\cdot7}$ which is true, but this incldues both the possibility of choosing $5$ and then $6$ and the possibility of choosing $6$ and then $5$.  When you then multiply by $4!$, I think you're doubling up.  I'm not $100\%$ certain though.
A: Re the glitch in your answer, (3) x (1/10)(1/9)(6/8)(5/7) x (4!)
(6/8)(5/7) implies that the two discs within the limiters are already permuted, so the (4!) needs to be replaced by (4!/2!)
If you had written (2/10)(1/9) for the limiters, the final factor would become (4!)/(2!2!)
It is because of such hidden snags that I prefer to use the safer route of combinations,  as @saulspatz has done.
