Alternate ways to teach this combinatorics question? Suppose we have a set A with 18 members. Further, A is divided into 3 subsets B, C, and D. Set B has 5 members, C has 6, and D has 7. If we randomly select 3 members of set A, what is the probability we'll select one member from each subset?
My question is this. What is the best way to solve this problem for high school students?
My method is this. First, find the probability that the first selection is from B, the second from C, and the third from D. This yields $\frac{35}{816}$. Since the order doesn't matter, and there are 3! = 6 possible orders, multiply by 6. Thus the probability is $\frac{35}{136}$.
This method works and is pretty easy. My students sometimes struggle to understand why I multiply by 6. I'd appreciate seeing any other solutions that might be a bit easier on the understanding, even if they require a bit more work.
 A: Another approach: Each of the $\binom{18}3$ sets of size 3 is equally likely to be chosen (by assumption). Among these, there are $5\cdot6\cdot7$ "good" sets (by the multiplication principle). So the probability of picking a good set is
$$\frac {5\cdot 6\cdot 7}{\binom{18}3} = \frac{35}{136}.$$
We don't multiply by $3!$ here, because the denominator doesn't count reordered versions of the same set as distinct, so we shouldn't count them separately in the numerator either. But if we want to do the multiple-counting in both places, that works too! We get
$$\frac {5\cdot 6\cdot 7\cdot3!}{P(18,3)}$$
where $P(18, 3)=18\cdot17\cdot16$ counts the partial permutations of 3 elements from an 18-element set. Reorderings are counted as distinct here, which is why we do the same in the numerator.
A student might question whether this version really models the given problem, in which order isn't supposed to matter—the idea is that we're imagining using an ordered process to select the unordered set (pick one element, then another, then another) and counting the ordered sequences that give us the desired unordered results. We should also justify why each possible ordered sequence is equally likely: of course, each one has probability $\frac1{18}\frac1{17}\frac1{16}$.
Alternatively, elaborating on your justification of your approach, I'd point out that $\frac{5}{18}\frac{6}{17}\frac{7}{16}$ is the probability of getting a member of $B$ first, a member of $C$ second, and a member of $D$ third in an ordered selection process. (Why? Because the ordering is what justifies those denominators: there are 17 elements remaining for the second selection after the first selection has been made, and so on. There's no "unordered version" of this calculation.) But a different "good" outcome is first getting a member of $C$, then a member of $B$, then a member of $D$; this happens with probability $\frac{6}{18}\frac{5}{17}\frac{7}{16}$. Considering all possible orders we could get, the probability of getting a good outcome is
$$
\frac{5}{18}\frac{6}{17}\frac{7}{16}+
\frac{5}{18}\frac{7}{17}\frac{6}{16}+
\frac{6}{18}\frac{5}{17}\frac{7}{16}+
\frac{6}{18}\frac{7}{17}\frac{5}{16}+
\frac{7}{18}\frac{5}{17}\frac{6}{16}+
\frac{7}{18}\frac{6}{17}\frac{5}{16}.
$$
(The addition here is justified since the six events are disjoint. They would only overlap if a single outcome represented multiple orderings of $B,C,D$, which is nonsensical.)
I imagine writing out the sum as above helps students see the point, and from there it's pretty easy see how it simplifies to
$$3!\frac{5}{18} \frac{6}{17} \frac{7}{16}.$$
