Differences between combinations and permutations. My problems are with the following question:

Four digits are selected independently at random (without repetition) from $\{0,1,...,9\}$. What is the probability that the four digits form a run? e.g $ (2,3,4,5)$

The probability of any selection is ${10 \choose 4}$ because we are not repeating digits. We can enumerate all the possible runs: $$\{0,1,2,3\},\{1,2,3,4\},...,\{6,7,8,9\}. $$ There are seven in total so the answer is $$7/{10 \choose 4}.$$
My query is as follows, the selection $\{1,2,3,4\} = \{2,1,3,4\}$, as we are dealing with combinations. But the r.h.s does not form a run. Does the question mean that it is possible that we can rearrange the 4 numbers into a run?
Or is it that by using combinations we enforce an ordering?
The answer agrees with what is said in the book. Should The answer be given in permutations instead?

The answer to your first question is positive, but not the second one. More specifically:

*

*Suppose the ordering does not matter. Then, the set of favorable outcomes will consist of 7 subsets as you correctly point out. The set of all outcomes will be comprised of C(10,4) subsets. So, the answer turns out to be 7/C(10,4).


*Now, suppose the ordering indeed matters. This time, the set of favorable outcomes will contain 7 permutations. On the other hand, the set of all outcomes will include C(10,4)*4!= P(10,4) permutations. Thus, the answer would be different: 7/P(10,4), or more equivalently, 7/C(10,4)*4!.
Hope this helps. Have a good day.
