Help needed with different approach to a combinatorics question Every day, the 15 students in Mr. Singh`s Advanced Chemistry class are randomly divided into 5 lab groups of 3 students each. What is the probability that three students - Linda, Martin, and Nancy - are in the same lab group today?
The answer to the question is to see it from the perspective of one of the students we are concerned with. From his/her point of view, 2 of the other 14 students are randomly chosen to be paired with her. This can be done in ${14}\choose{2}$ = 91 ways. Only one of those pairs are her "friends", so the answer is $\frac{1}{91}$.
This answer of course didn't come to me, I read it when I couldn't figure it out. Though curious if I could find the same answer with my approach, I tried to kind of work backwards, knowing the answer.
Here's what I tried to do: I instead thought of how many groups of three could be chosen from 15 students. I thought this could be done in ${15}\choose{3}$ = 455 ways. One of those contains the group of students we are interested in. But the answer is not $\frac{1}{455}$. How does 91 fit into this? Curious as I was, I divided 455 by 91, and lo and behold, I got 5. Did this 5 mean the 5 groups of three students? I wasn't sure.
If I divide 455 by 5, I get 91. But what is the meaning of this? I would rather think that, from the 455 available choices, we choose 5 things (in this case a thing is a group of 3 students). But this would mean ${455}\choose{5}$ = some huge number.
Let's say I have ABCDE, and I want to choose 2 of them. This would be ${5}\choose{2}$ = 10 combinations. If I would ask "how many groups of 5 can I make?", then the answer surely isn't $\frac{10}{5}$ = 2. I definitely can check with pen and paper and see that I can make quite more than 2 groups. So I know my thinking is faulty.
But where? What does the 5 in $\frac{455}{5}$ = 91 mean? And what should my line of thought be when starting from ${15}\choose{3}$ to get to the answer?

There are $\binom{15}3=455$ possible $3$-person groups, so if we were picking just one group of $3$ at random, the probability would indeed be $\frac1{455}$ that we picked this specific group. However, we’re not just picking a single group: we’re dividing the $15$ students into five groups of $3$, so we actually have $5$ chances to get the desired group, not just one. And $\frac5{455}$ is, as you noticed, $\frac1{91}$.
You might reasonably worry that picking a random partition of the students into $5$ groups of $3$ students isn’t really the same as picking $5$ groups of $3$ at random with replacement from the collection of all $455$ $3$-person groups. We can do a more elaborate calculation to show that it really does work out right. There are
$$\binom{15}3\binom{12}3\binom93\binom63\binom33$$
ways to pick a first group of $3$, then pick a second group of $3$ from the $12$ people who remain, and so on until we have all $5$ groups. However, this counts each of the $5!$ permutations of the $5$ groups separately, so the number of ways to partition the students into $5$ groups of $3$ is really only
$$\frac1{5!}\binom{15}3\binom{12}3\binom93\binom63\binom33\;.$$
How many of these partitions have the desired trio as one of its parts? The same reasoning shows that if we set Linda, Martin, and Nancy aside and form $4$ groups of $3$ from the remaining $12$ students, we can do this in
$$\frac1{4!}\binom{12}3\binom93\binom63\binom33$$
different ways. The probability that a randomly chosen partition has our trio as one of its groups is therefore
$$\frac{\frac1{4!}\binom{12}3\binom93\binom63\binom33}{\frac1{5!}\binom{15}3\binom{12}3\binom93\binom63\binom33}=\frac{5!}{4!\binom{15}3}=\frac5{\binom{15}3}=\frac5{455}=\frac1{91}\;.$$
