Expected number of boy-girl pairs $4n$ children at a party are paired at random, with each pair being equally likely. If there are $n$ boys and $3n$ girls, find the expected number of boy-girl pairs. (Ordering does not matter within boy-girl pairs or between pairs)
So far I've attempted:
Let $x$ be the number of pairs consisting of a boy and a girl.
Possible values of $x$ are from $0$ to $n$.
$E(x)=∑_{i,j=0}^{n} P(x_{i,j})$
where $x_{i,j}$ is an indicator random variable that is equal to $1$ if boy $i$ is paired with girl $j$, and 0 otherwise.
However, I'm not sure how to calculate $P(x_{i,j})$
My guess is that it would be $\frac{\binom{n}{1}\cdot\binom{3n}{1}}{\binom{4n}{2}}$, but I'm not sure if this is overcounting.
Also, after we find $P(x_{i,j})$, do we sum $P(x_{i,j})$ over $n$ possible pairs to find $E(x)$?
 A: You ask specifically about your probability $P(X_{i,j})$ the probability that boy $i$ is paired with girl $j$.
The specific boy, boy $i$, will be in one of the pairs.  It matters not which to us.  The partner paired with boy $i$ is equally likely to be any of the other $4n-1$ children, exactly one of which is the specific girl, girl $j$.  The probability then you ask about is simply $$P(X_{i,j})=\dfrac{1}{4n-1}$$
"After we find $P(X_{i,j})$ do we sum $P(X_{i,j})$ over $n$ possible pairs to find $E[X]$?"  No, we are summing over all possible boy-girl pairings.  There are $n\times (3n)$ possible pairings, namely boy 1 with girl 1, boy 1 with girl 2, boy 1 with girl 3, ... boy n with girl 3n-1, boy n with girl 3n.
We can see that $X = \sum\limits_{i=1}^n\sum\limits_{j=1}^{3n} X_{i,j}$ and continue from there with linearity of expectation.

An alternate approach would have been instead of looking at each of the possible pairings (of which there are $3n^2$) to look at each of the pairs (of which there are only $2n$).
(Note, with $4n$ children, there are $2n$ pairs made... not just $n$)
Letting $Y_i$ be the indicator random variable which equals $1$ if the $i$'th pair has one boy and one girl and $0$ otherwise, there are $\binom{4n}{2}$ equally likely pairs of children which could be in this $i$'th pair, $n\times 3n$ of which are a boy-girl pairing.  This gives $\Pr(Y_i) = \dfrac{n\times 3n}{\binom{4n}{2}}$.
We can then recognize that $X = \sum\limits_{i=1}^{2n} Y_i$ and continue from there with linearity of expectation.

We could have broken up yet another way... letting $Z_i$ be the indicator random variable corresponding to whether or not boy $i$  was partnered with a girl.  You'd have $X = \sum\limits_{i=1}^n Z_i$ and you'd have $P(Z_i)=\frac{3n}{4n-1}$.  Similarly you could have done this from the girls' perspectives.


 We have from the first, $E[X] = E[\sum\limits_{i=1}^n\sum\limits_{j=1}^{3n} X_{i,j}] = n\times 3n \times \dfrac{1}{4n-1}$.  We have from the second $E[X] = E[\sum\limits_{i=1}^{2n}Y_i] = 2n\times \dfrac{n\times 3n}{\binom{4n}{2}}$.  From the third we have $E[X]=E[\sum\limits_{i=1}^n Z_i] = n\times \dfrac{3n}{4n-1}$.  You should be able to see after a little algebraic manipulation that these are all of course equal.

A: The probability that any given boy is paired with a girl is $\frac{3n}{4n-1}$.  By linearity of expectation (and since there are $n$ boys), the expected number of boy-girl pairs is $n\cdot\frac{3n}{4n-1}=\frac{3n^2}{4n-1}$.
A: You are on the right track.
This is a finite problem, so the expected value is taken over all the ways to select $2n$ children pairs from the $4n$ children. How many ways are there? Let's do it generally for $k$ pairs of $2k$ children (using numbers from $1$ to $2k$ to represent the children):
You can transform the pairs into a permuation of the $2k$ children by writing down the 2 children in one pair as first and second entry of the permutation, then 2 children of another pair as 3rd and 4th, a.s.o.
For $k=3$, the pairing $(1,4),(5,2),(3,6)$ could be written down as $145236$ or $412536$ or $523614$ and many more ways.
So how many of the $(2k)!$ permutations correspond to the same pairings arrangement? Well, inside each pair the order of the children doesn't matter (compare $145236$ and $412536$ above). Each pair can be ordered in $2!=2$ ways, since we have $k$ pairs this reduces the number pairings by a factor or $2^k$.
In addition the order we look at the $k$ pairs doesn't matter either (compare $145236$ and $523614$ above), this reduces the number of pairings by an additional factor of $k!$.
But that's all the symmetries for the pairings $\to$ permutations algorithm above. If we fix the order of the pairs and the order inside each pair, we now get exactly one permutation from the pairing.
So, the number of pairings for $2k$ children into $k$ pairs is
$$P_{2k} = \frac{(2k)!}{2^kk!}.$$
So what is now the expected value of your indicator random variable $x_{i,j}$, that the pair $(i,j)$ is part of the pairing? We know that there are $P_{4n}$ equiprobable pairings. For each such pairing the pair $(i,j)$ is either in it exactly once or not at all. So in how many pairings is it?
Well, if you have already "preselected" this pair into the pairing, you now have to make $2n-1$ pairs from the remaining $4n-2$ children, which there are $P_{4n-2}$ ways of.
That means
$$E(x_{i,j}) = \frac{P_{4n-2}}{P_{4n}} = \frac{(4n-2)!}{2^{2n-1}(2n-1)!} \frac{2^{2n}(2n)!}{(4n)!} = \frac{2\cdot 2n}{(4n)(4n-1)} = \frac1{4n-1}.$$
But since we are interested in boy-girl pairs in general and not a specific one, we need to multiply the above by the number of boy-girl pairs, which is simply $n\cdot 3n =3n^2$.
That means the final result for the expected value of the number of boy-girl pairs is
$$E_n=\frac{3n^2}{4n-1}.$$
It's always good to check the forumla for small values where the result can be found in other ways, to ward against errors during calculations.
For $n=1$, each pairing has to pair the single boy with a girl, so each pairing contains exacly one boy-girl pair, so the expected values is $1$ as well, which the above formula correctly yields.
For $n=2$, a specific boy ($B_1$) can be paired with the seven other childs in $7$ ways: With the other boy $B_2$ or one of the 6 girls $(G_1,\ldots, G_6)$, with equal probability.
In any pairings with where $B_1$ is paired with $B_2$, there is no boy-girl pair (probability: $\frac17$). In any pairings where $B_1$ is paired with a girl, there are exactly 2 boy-girl pairs (probability: $\frac67$). So the expected value of boy-girl pairs for $n=2$ is
$$\frac17\cdot 0 + \frac67\cdot2 = \frac{12}7,$$
which is again what the above formula predicts.
