Combinatorial solution: What is the probability that exactly two friends are in the same class? Consider three friends (one of which is called Jon) in a group of 90 students. The professor splits the group into three classes of equal size. 
a) What is the probability that exactly two friends are in the same class?
b) What is the probability that Jon ends up together with exactly one of his friends in a class?
My Solution: a) Let's say we put one of the friends in a class at random. Then there is $\frac{29}{89}$ chance for the second friend to be in the same class as the first and a $\frac{60}{88}$ chance for the last friend to be in a different class. There are three ways of arranging the friends this way, thus the probability is $3\frac{29}{89}\frac{60}{88}$.
b) Because Jon is now fixed, there are only two ways to achieve the desired outcome, so the probability is $2\frac{29}{89}\frac{60}{88}$.
I have two questions:


*

*Are these results correct?

*Is there a way to get the solutions in a combinatorial way? 
 A: Both of your solutions are correct.
For a combinatorial approach, we will treat the groups as unlabeled.
There are 
$$\binom{90}{30}\binom{60}{30}\binom{30}{30}$$
to select three labeled groups of $30$ people each from a group of $90$ people. Since the order in which the groups are selected does not matter, the number of ways of selecting three unlabeled groups of $30$ people each from a group of $90$ people is
$$\frac{1}{3!}\binom{90}{30}\binom{60}{30}\binom{30}{30}$$
To count the number of ways exactly two of the friends are placed in the same class, choose which two of the three friends are placed together, which $28$ of the other $87$ people are placed in their group, and which $29$ of the remaining $59$ people are placed in the same group as the other friend.  This gives
$$\binom{3}{2}\binom{87}{28}\binom{1}{1}\binom{59}{29}\binom{30}{30}$$
favorable cases.
Therefore, the probability that exactly two of the three friends are placed in the same class is 
$$\frac{\dbinom{3}{2}\dbinom{87}{28}\dbinom{1}{1}\dbinom{59}{29}\dbinom{30}{30}}{\dfrac{1}{3!}\dbinom{90}{30}\dbinom{60}{30}\dbinom{30}{30}}$$
To count the number of ways Jon is placed with exactly one of his friends, choose which friend is placed with Jon, which $28$ of the remaining $87$ people are placed in their group, and which $29$ of the remaining $59$ other people are placed in the group with their friend.  This gives
$$\binom{2}{1}\binom{87}{28}\binom{1}{1}\binom{59}{29}\binom{30}{30}$$
favorable cases.
Therefore, the probability that Jon is placed in the same class as exactly one of his friends is 
$$\frac{\dbinom{2}{1}\dbinom{87}{28}\dbinom{1}{1}\dbinom{59}{29}\dbinom{30}{30}}{\dfrac{1}{3!}\dbinom{90}{30}\dbinom{60}{30}\dbinom{30}{30}}$$
