"Six brothers sign up to take a class. The class has 90 registered students including the six. There are three sections, A, B, and C, of 30 students each by random assignment. What is the probability that each class section has exactly 2 brothers?"
What I've tried: I know there are 15 different ways to pair the brothers, from 6 factorial divided by (3 factorial times (2 factorial) cubed)). But I don't know how to set up the probability of any given pairing scheme seeing pairs ending up together, for all three classes.
I figured we could let the first brother go anywhere, then his partner had a 29/89 chance of ending up in his section, then the next brother had to go to either remaining section (60/88), his partner had to follow (29/87), then the last two take up the last section (30/86 * 29/85).
Am I overcounting or undercounting due to ordering? What is the above calculation finding the probability of, if it is not the probability of one pairing scheme ending up in separate classes?
Thanks.
There are ${90}\choose{30}$ ways to choose section A, then based on that selection there are ${60}\choose{30}$ ways to choose section B, then finally ${{30}\choose{30}} = 1$ way to choose section C, so the product of those is the number of ways to arrange the students.
Similarly, there are ${6}\choose{2}$ ways to choose two brothers to go in section A, and ${84}\choose{28}$ ways to choose the rest of the students. Can you see what the expressions will be for the other two sections, and hence an expression for the number of ways to put two brothers into each class?
Then, based on that, can you work out what the probability is?