I'm confused whether to use permutation or combination in the question given below?

Question

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 $7$?

In the above question I have doubt in the answer given, I'm giving the link to the accepted answer here: I can't figure out this combinatorics problem... Or at least why my solution doesn't work.

In the solutions given in the link they took combination i.e. they took $^6C_2$ for the other two numbers, shouldn't it be a permutation, because it is given in the question that the disks are drawn one after another without replacement and since one after the other is given the order matters. I'm thinking of it this way if I have to fill two empty chairs from $6$ people one after the other, the first chair has $6$ possibilities and the second one has $5$ possibilities, so $30$ in total so basically I'm doing permutation because order matters since one by one is mentioned in the question.

Answer 1

I'm going to rephrase the problem into 2 similar ones in order to attempt to make you see a difference.

First

There are 10 labeled disks in a box and we are drawing 7 disks one at a time without replacement. We want to determine the probability that the range of numbers of the selected disks is 7.

Second

There are 10 labeled disks in a box and we are drawing 7 disks one at a time without replacement. Now we want to only keep every second disk and the others we return back to the box. The disks that we keep are considered selected. What is the probability that the range of numbers of the selected disks is 7?

Now let's consider the draws: $ 1\ 4\ 6\ 9\ 2\ 3\ 5\ $ and $6\ 5\ 2\ 1\ 3\ 9\ 4\ $

They contain the same set of numbers.

In the first problem we get to keep all the drawn discs and as you can see the range of numbers is not changed regardless of the order we draw the discs.

However in the second problem we get to keep only discs $4\ 9\ 3\ $ from the first draw and discs labeled $5\ 1\ 9$ from the second. We can see that the order in which the disks are drawn in the second problem actually influences the probability. We took the same set of numbers and ended up with a different range.

Answer 2

You can solve this problem using permutations if you want. If you are solving it using permutations, then there are $10 \cdot 9 \cdot 8 \cdot 7 = 5040$ possible selections. Since you want a range of $7$, then each selection must include at least $\{1,8\}$, $\{2,9\}$, $\{3,10\}$ and be bounded by those values so that the range does not exceed $7$. Then you have $12$ possibilities for each set of numbers:

$$ 10 \space 3 \space \_ \space \_ \\ 10 \space \_ \space 3 \space \_ \\ 10 \space \_ \space \_ \space 3 \\ \_ \space 10 \space 3 \space \_ \\ \_ \space 10 \space \_ \space 3 \\ \_ \space \_ \space 10 \space 3 $$

$$ 3 \space 10 \space \_ \space \_ \\ 3 \space \_ \space 10 \space \_ \\ 3 \space \_ \space \_ \space 10 \\ \_ \space 3 \space 10 \space \_ \\ \_ \space 3 \space \_ \space 10 \\ \_ \space \_ \space 3 \space 10 $$

Each of thoses cases above can have $6*5 = 30$ possibilities since the other two numbers are chosen without replacement and must come from $\{4,5,6,7,8,9\}$ in this specific example. So we have a total of $30*12 = 360$ possibilities. Repeat this process for $\{1,8\}$ and $\{2,9\}$ and you get a total of $360*3 = 1080$ possibilities.

$1080$ selections where the range is $7$ out of $5040$ total selections means there is a $1080/5040 = 3/14$ probability of selecting a set of disks whose range is 7.