Combinatorics problem: Splitting 90 students into 3 classes A group of 90 students is to be split at random into 3 classes of equal size. All partitions are equally likely. Joe and Jane are members of the 90-student group. Find the probability that Joe and Jane end up in the same class.
Answer:


Another way of thinking of this:
The different ways of partitioning into 3 partitions, each of size 30 is:
all = 90! / (30!)^3
The number of ways in which Joe and Jane can be at the same class is:
x = 88! / (30! * 30! * 28!)
and since there are 3 classes, the answer is 3 times the above expression:
y = 3 * x
Final answer:
y / all
This way of thinking works.

Yet, another way:
We can consider both Joe and Jane to be 1 student, in other words, think of them as one unit, and consider the number of students to be 89, and the partitions to be 30, 30, and 29.
This way, the number of ways in which they would be together:
z = 89! / (30! * 30! * 29!)
and the final answer would be z / all.
Unfortunately, this is not true.
you can see that z = x * 89/29 = x * 3.06896551724 which is not equal to x * 3.
What's wrong with the last logic?
 A: What is wrong with the last approach is that not all configurations are equally probable.  As we see in the others, the chance is less than $1/3$ because the first one you put in a class takes up one space, so there are fewer spaces left for the other.
A: 29/89 is the correct answer.  However the analysis is overly complicated.  Simple approach:  There are 29 students in the same class as Joe and 60 in the other classes.  Since Jane's class assignment is random, she has 29/89 chance of being in the same class as Joe.
A: Whatever class Joe is in, there are 29 other students.  Jane can be selected to be one among these with a probability of $29/89$.
This may also be viewed as: there are $\binom 31\binom{30}2$ ways to select two seats in the same class, and $\binom{90}{2}$ ways to select two seats from anywhere among the three classes. So the probability for obtaining seats in the same class when selecting two from the ninety seats, without bias, for Joe and Jane is: $$\dfrac{3\cdot 30\cdot 29/2}{90\cdot 89/2}$$


What is wrong with the last logic?

We can consider both Joe and Jane to be 1 student,


No.  We cannot.  If they were super-glued together, they would certainly be in the same class.
