90 students, including Vivien and Victoria, are to split into three classes of equal size, and this is to be done at random. What is the probabilty that Vivien and Victoria end up in the same class?
My attempt is $\frac{\binom {88}{28}\binom {60}{30}\binom {30}{30}}{\binom {90}{30}\binom {60}{30}\binom {30}{30}}$ but the correct answer is 3 times my answer.
I thought that all three classes are the same because they have same amount of people?
The simplest way is to seat Vivien and Victoria first. Vivien can be seated anywhere, and Victoria must occupy one of the $29$ remaining seats in that class out of 89 total remaining,
Thus $ Pr =\frac{29}{89}$
In short, the classes have the same amount of students but ultimately are not the same.
To see this, let us look at the case where the classes do not have the same amount of students, say, 5, 10, and 15.
Then the answer will be \begin{equation} \frac{\binom{28}{3}\binom{25}{10}\binom{15}{15} + \binom{28}{8}\binom{20}{15}\binom{5}{5} + \binom{28}{13}\binom{15}{5}\binom{10}{10}}{\binom{30}{5}\binom{25}{10}\binom{15}{15}} \end{equation}
As you can see, there are three terms in the numerator corresponding to the class Vivien and Victoria are in. The case where the classes have the same number of students is no different, these terms are just equal to each other.
This is the intuitive way to see what is wrong in your argument. Here is the rigorous proof.
To strictly define the random process of splitting people into three groups we need to state how to calculate the probability. In your problem it is done in the following way. Let $\Omega$ denote the set of all possible outcomes (i.e. of all the triples of classes). Let $X\subset\Omega$ denote the set of outcomes where Vivien and Victoria are in the same class. Then the desired probability is equal to $|X|/|\Omega|$.
Notice that here the classes can be ordered, meaning that if we swap any two classes, we get a different outcome, or unordered, meaning that if we can swap any two classes and get the same outcome. Here might lie the source of your confusion.
In the first case we have $|\Omega| = \binom{90}{30}\binom{60}{30}\binom{30}{30}$, but we have $|X| = 3\binom{88}{28}\binom{60}{30}\binom{30}{30}$ because we "care" which class Vivien and Victoria are in. Thus the numerator ($|X|$) is three times larger than your numerator.
In the second case $|X|$ is $2! = 2$ times smaller than your numerator because we may assume that Vivien and Victoria are in the first class but the other two classes may go in any order. However, in this case $|\Omega|$ is $3! = 6$ times smaller than your denominator (because we essentially identify 6 outcomes from the set of permutations of the three classes), so the answer is once again 3 times larger than yours.
Given any class of $30$, the probability that (if) Vivien is in the class, that even Victoria is in it = $P(Vr|Vn) =\frac{1}{30} \times \frac{1}{29}$
The total number of classes of $30$ = $^{90}C_{30} + ^{60}C_{30} + ^{30}C_{30}$
Is this sufficent information to find the answer?
