I answered this question correctly using combinations rather than permutations. Is my solution valid, or am I misreading the question?
"A sample of size 3 is drawn at random and without replacement from the population $\{1, 2, 3, 4, 5\}$. What is the probability that the range of the sample is equal to 3?"
My logic: Because we are interested in the events where the range is three, I concluded that the cardinality of the sample N from the population $P$ is #$(N) = 4$ because $N = \{(1, 2, 4), (1, 3, 4), (2, 3, 5), (2, 4, 5)\}$.
So then the $P($range$ = 3) =\dfrac{4}{5 \choose 3} = \dfrac{4}{10}$.
Solution: There are six possible orderings for both $(1,2,4)$ and $(1,3,4)$, which is also true for both $(2,3,5)$ and $(2,4,5)$. So there are 24 ways to select the three good sets and $(5)(4)(3)= 60$ ways to select three from $P$. So $\dfrac{24}{60} = \dfrac{4}{10}$
Does the phrase "without replacement" imply one at a time? Or is there something in the phrasing of this question that should signal to me to use permutations rather than combinations? I was under the impression that the order they are chosen doesn't matter because the range relies on the two numbers being chosen in such a way that they are the max and minimum and the difference is 3.
Your solution is valid.
You counted ways to obtain a combination with a range of 3 (maximum-minimum) when selecting three numbers. [Select a minumum, select a middle, select a maximum such that maximum-minum=3 vs. select any three from the five numbers.] $$\dfrac{{^2\mathrm C_1}{^2\mathrm C_1}{^1\mathrm C_1}}{^5\mathrm C_3}$$
They counted the way to obtain a permutation with a range of 3 (maximum-minimum) when selecting and arranging three numbers.
$$\dfrac{{^2\mathrm C_1}{^2\mathrm C_1}{^1\mathrm C_1}~ 3!}{^5\mathrm C_3~ 3!}$$
Well, now you see that here, as often is the case, the extra effort of counting ways to arrange is redundant because the common factor cancels.
Remark: Generally when order is important (for the event), then the probability is a ratio of permutation to combination. However, when order is not important (for the event), then the probability may equally be either a ratio of combinations, or a ratio of permutations. (Assuming all atomic outcomes being counted are equally probable)