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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.

Is my solution valid, or am I misreading the question?

Your solution is valid.

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.

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.]
$$dfrac{{^2mathrm C_1}{^2mathrm C_1}{^1mathrm C_1}}{^5mathrm C_3}$$

They counted the way to obtain a permutation with a range of 3 (maximum-minimum) when selecting and arranging three numbers.

$$dfrac{{^2mathrm C_1}{^2mathrm C_1}{^1mathrm C_1}~ 3!}{^5mathrm C_3~ 3!}$$

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)