# Dividing a class into two equal groups , I can't find my mistake

Suppose we have know A and B are two students in a class of $$2n$$ pupils. We divide the class into two groups of n each. What is the probability that they will be in the same group?

I calculate the complement, that is, the probability that they are in different groups. As they are in different groups, I need to select $$n-1$$ students from $$2n-2$$.

Thus, my answer is $$1 - \frac{2n-2 \choose n-1}{2n \choose n}$$, which ends up differing from the answer given at Probability of getting into the same class.

Can somebody help by explaining the mistake I have made?

Please note that $${2n \choose n}$$ orders the groups so you need to divide it by $$2!$$ as you only want to find number of ways to make two groups of $$n$$.

So the probability using your method is,

$$\displaystyle 1 - 2! \cdot {2n-2 \choose n-1} / {2n \choose n} = \frac{n-1}{2n-1}$$

The easiest way to find the probability is to first place $$A$$ in any group. Now that leaves $$(2n-1)$$ places across two groups for $$B$$ but if $$B$$ has to be in the same group as $$A$$ then it must be in one of the $$(n-1)$$ places in $$A$$'s group. That is probability of $$\displaystyle \frac{n-1}{2n-1}$$.

A bit more explanation:

Say I have $$4$$ guests at home but I have only two chairs. Say I want to randomly choose two guests to sit and of course rest two stand along with me. $${4 \choose 2}$$ gives me number of ways to make two groups. But these groups are ordered. One of the groups sits and the other group stands.

But if I am only interested in finding number of ways I can randomly pair my $$4$$ guests to play a doubles game of ping pong, there are only $$\frac{1}{2!}{4 \choose 2} ~$$ ways. By the way, my four guests are Alpha, Bravo, Charlie and Delta. Here are three ways to pair them -

• $$\{$$Alpha, Bravo$$\}$$,$$\{$$Charlie, Delta$$\}$$

• $$\{$$Alpha, Charlie$$\}$$,$$\{$$Bravo, Delta$$\}$$

• $$\{$$Alpha, Delta$$\}$$,$$\{$$Bravo, Charlie$$\}$$

So for the given question to find desired probability, you either need to have number of unordered groups in both the numerator and the denominator or number of ordered groups in both. To make the numerator ordered, note that $$A$$ and $$B$$ can swap places to be in named groups and so you multiply $${2n-2 \choose n-1}$$ by $$2!$$.

• Where can I read more about the fact that $2n \choose n$ orders the groups? I understand your point here, but I'm not sure if there are some other things that might be unclear in my foundations. Dec 5 '21 at 17:43
• Dec 5 '21 at 17:49