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


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


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


different ways. The probability that a randomly chosen partition has our trio as one of its groups is therefore


  • $\begingroup$ "5 chances to get the desired group" - I try to think about this, but I'm having a hard time. I keep connecting the 5 groups with the 455 things, which I probably shouldn't do. As an aside, how long did it take you to be fluent in this, i.e., have the insight quick and the ability to see it from multiple angles? $\endgroup$ – Garth Marenghi Oct 20 '16 at 19:37
  • $\begingroup$ @Garth: When you divide the students into $5$ groups of $3$, you really are picking $5$ groups, not just one, and one of those $5$ could be the one that we want. I really can’t answer that last question: it’s been over $50$ years since I first encountered these ideas. I don’t think that I ever had any trouble with them, but I know from having taught the subject that my experience is not typical. $\endgroup$ – Brian M. Scott Oct 20 '16 at 19:46

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